{"id":14807,"date":"2018-09-27T18:36:30","date_gmt":"2018-09-27T18:36:30","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/series-and-their-notations-2\/"},"modified":"2025-07-29T13:59:00","modified_gmt":"2025-07-29T13:59:00","slug":"series-and-their-notations-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/series-and-their-notations-2\/","title":{"raw":"Series and Their Notations","rendered":"Series and Their Notations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Use summation notation.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use the formula for the sum of the \ufb01rst n terms of an arithmetic series.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use the formula for the sum of the \ufb01rst n terms of a geometric series.<\/li>\r\n \t<li style=\"font-weight: 400;\">Use the formula for the sum of an in\ufb01nite geometric series.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve word problems involving series.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Using Summation Notation<\/h2>\r\nTo find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a <strong>series<\/strong>. Consider, for example, the following series.\r\n<div style=\"text-align: center;\">[latex]3+7+11+15+19+\\dots[\/latex]<\/div>\r\nThe <strong> [latex]n\\text{th }[\/latex] partial sum<\/strong> of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation [latex]\\text{ }{S}_{n}\\text{ }[\/latex] represents the partial sum.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{S}_{1}=3\\\\ &amp;{S}_{2}=3+7=10\\\\ &amp;{S}_{3}=3+7+11=21\\\\ &amp;{S}_{4}=3+7+11+15=36\\end{align}[\/latex]<\/div>\r\n<strong>Summation notation <\/strong>is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>, [latex]\\sigma[\/latex], to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term in the series. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term in a series.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183627\/CNX_Precalc_Figure_11_04_001n2.jpg\" alt=\"Explanation of summation notion as described in the text.\" \/>\r\n\r\nIf we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{k}=2k[\/latex] for [latex]k=1[\/latex] through [latex]k=5[\/latex]. We can begin by substituting the terms for [latex]k[\/latex] and listing out the terms of this series.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {a}_{1}=2\\left(1\\right)=2 \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/div>\r\nWe can find the sum of the series by adding the terms:\r\n<div style=\"text-align: center;\">[latex]\\sum _{k=1}^{5}2k=2+4+6+8+10=30[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Summation Notation<\/h3>\r\nThe sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:\r\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{n}{a}_{k}[\/latex]<\/p>\r\nThis notation tells us to find the sum of [latex]{a}_{k}[\/latex] from [latex]k=1[\/latex] to [latex]k=n[\/latex].\r\n\r\n[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<h3>Does the lower limit of summation have to be 1?<\/h3>\r\n<em>No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given summation notation for a series, evaluate the value.<\/h3>\r\n<ol>\r\n \t<li>Identify the lower limit of summation.<\/li>\r\n \t<li>Identify the upper limit of summation.<\/li>\r\n \t<li>Substitute each value of [latex]k[\/latex] from the lower limit to the upper limit into the formula.<\/li>\r\n \t<li>Add to find the sum.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Using Summation Notation<\/h3>\r\nEvaluate [latex]\\sum _{k=3}^{7}{k}^{2}[\/latex].\r\n\r\n[reveal-answer q=\"14937\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"14937\"]\r\n\r\nAccording to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{k}^{2}[\/latex] from [latex]k=3[\/latex] to [latex]k=7[\/latex]. We find the terms of the series by substituting [latex]k=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{k}^{2}[\/latex]. We add the terms to find the sum.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sum _{k=3}^{7}{k}^{2}&amp; ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill &amp; =9+16+25+36+49\\hfill \\\\ \\hfill &amp; =135\\hfill \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]\\sum _{k=2}^{5}\\left(3k - 1\\right)[\/latex].\r\n\r\n[reveal-answer q=\"812548\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"812548\"]\r\n\r\n38\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172726[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Using the Formula for Arithmetic Series<\/h2>\r\nJust as we studied special types of sequences, we will look at special types of series. Recall that an <strong>arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the <strong>common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an <strong>arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:\r\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/div>\r\nWe can also reverse the order of the terms and write the sum as\r\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/div>\r\nIf we add these two expressions for the sum of the first [latex]n[\/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[\/latex] terms of any arithmetic series.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}\\hfill \\\\ +{S}_{n}&amp;={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}\\\\ \\hline 2{S}_{n}&amp;=\\left({a}_{1}+{a}_{n}\\right)+\\left({a}_{1}+{a}_{n}\\right)+...+\\left({a}_{1}+{a}_{n}\\right)\\end{align}[\/latex]<\/div>\r\nBecause there are [latex]n[\/latex] terms in the series, we can simplify this sum to\r\n<div style=\"text-align: center;\">[latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/div>\r\nWe divide by 2 to find the formula for the sum of the first [latex]n[\/latex] terms of an arithmetic series.\r\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of an Arithmetic Series<\/h3>\r\nAn <strong>arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the sum of the first [latex]n[\/latex] terms of an arithmetic sequence is\r\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given terms of an arithmetic series, find the sum of the first [latex]n[\/latex] terms.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\r\n \t<li>Determine [latex]n[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1}\\text{, }{a}_{n}[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Finding the First <em>n<\/em> Terms of an Arithmetic Series<\/h3>\r\nFind the sum of each arithmetic series.\r\n<ol>\r\n \t<li>[latex]5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32[\/latex]<\/li>\r\n \t<li>[latex]20 + 15 + 10 +\\dots + -50[\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"535733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"535733\"]\r\n<ol>\r\n \t<li>We are given [latex]{a}_{1}=5[\/latex] and [latex]{a}_{n}=32[\/latex].Count the number of terms in the sequence to find [latex]n=10[\/latex]. Substitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{10}&amp;=\\frac{10\\left(5+32\\right)}{2}=185 \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>We are given [latex]{a}_{1}=20[\/latex] and [latex]{a}_{n}=-50[\/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{n}&amp;={a}_{1}+\\left(n - 1\\right)d \\\\ -50&amp;=20+\\left(n - 1\\right)\\left(-5\\right)\\\\ -70&amp;=\\left(n - 1\\right)\\left(-5\\right) \\\\ 14&amp;=n - 1\\\\ 15&amp;=n\\end{align}[\/latex]<\/div>\r\nSubstitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align} {S}_{n}&amp;=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{15}&amp;=\\frac{15\\left(20 - 50\\right)}{2}=-225\\end{align}[\/latex]<\/div><\/li>\r\n \t<li>To find [latex]{a}_{1}[\/latex], substitute [latex]k=1[\/latex] into the given explicit formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{k}&amp;=3k - 8 \\\\ {a}_{1}&amp;=3\\left(1\\right)-8=-5\\end{align}[\/latex]<\/div>\r\nWe are given that [latex]n=12[\/latex]. To find [latex]{a}_{12}[\/latex], substitute [latex]k=12[\/latex] into the given explicit formula.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{k}&amp;=3k - 8 \\\\ {a}_{12}&amp;=3\\left(12\\right)-8=28 \\end{align}[\/latex]<\/div>\r\nSubstitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{12}&amp;=\\frac{12\\left(-5+28\\right)}{2}=138 \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nUse the formula to find the sum of each arithmetic series.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\text{1}\\text{.4 + 1}\\text{.6 + 1}\\text{.8 + 2}\\text{.0 + 2}\\text{.2 + 2}\\text{.4 + 2}\\text{.6 + 2}\\text{.8 + 3}\\text{.0 + 3}\\text{.2 + 3}\\text{.4}[\/latex]\r\n\r\n[reveal-answer q=\"715922\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"715922\"]\r\n\r\n26.4\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]\r\n\r\n[reveal-answer q=\"81811\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"81811\"]\r\n\r\n328\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\sum _{k=1}^{10}5 - 6k[\/latex]\r\n\r\n[reveal-answer q=\"16304\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16304\"]\r\n\r\n-280\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]23739[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Solving Application Problems with Arithmetic Series<\/h3>\r\nOn the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?\r\n\r\n[reveal-answer q=\"924199\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924199\"]\r\n\r\nThis problem can be modeled by an arithmetic series with [latex]{a}_{1}=\\frac{1}{2}[\/latex] and [latex]d=\\frac{1}{4}[\/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[\/latex], and we are looking for [latex]{S}_{8}[\/latex]. To find [latex]{a}_{8}[\/latex], we can use the explicit formula for an arithmetic sequence.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{n}&amp;={a}_{1}+d\\left(n - 1\\right) \\\\ {a}_{8}&amp;=\\frac{1}{2}+\\frac{1}{4}\\left(8 - 1\\right)=\\frac{9}{4} \\end{align}[\/latex]<\/p>\r\nWe can now use the formula for arithmetic series.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} {S}_{n}&amp;=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{8}&amp;=\\frac{8\\left(\\frac{1}{2}+\\frac{9}{4}\\right)}{2}=11 \\end{align}[\/latex]<\/p>\r\nShe will have walked a total of 11 miles.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nA man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?\r\n\r\n[reveal-answer q=\"505026\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"505026\"]\r\n\r\n$2,025\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using the Formula for Geometric Series<\/h2>\r\nJust as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>. Recall that a <strong>geometric sequence<\/strong> is a sequence in which the ratio of any two consecutive terms is the <strong>common ratio<\/strong>, [latex]r[\/latex]. We can write the sum of the first [latex]n[\/latex] terms of a geometric series as\r\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[\/latex].<\/div>\r\nJust as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first [latex]n[\/latex] terms of a geometric series. We will begin by multiplying both sides of the equation by [latex]r[\/latex].\r\n<div style=\"text-align: center;\">[latex]r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}[\/latex]<\/div>\r\nNext, we subtract this equation from the original equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1} \\\\ -r{S}_{n}&amp;=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right) \\\\ \\hline\\left(1-r\\right){S}_{n}&amp;={a}_{1}-{r}^{n}{a}_{1}\\end{align}[\/latex]<\/div>\r\nNotice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for [latex]{S}_{n}[\/latex], divide both sides by [latex]\\left(1-r\\right)[\/latex].\r\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of a Geometric Series<\/h3>\r\nA <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence. The formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as\r\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{, r}\\ne \\text{1}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a geometric series, find the sum of the first <em>n<\/em> terms.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Finding the First <em>n<\/em> Terms of a Geometric Series<\/h3>\r\nUse the formula to find the indicated partial sum of each geometric series.\r\n<ol>\r\n \t<li>[latex]{S}_{11}[\/latex] for the series [latex]\\text{ 8 + -4 + 2 + }\\dots [\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{6}3\\cdot {2}^{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"792584\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"792584\"]\r\n<ol>\r\n \t<li>[latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex].We can find [latex]r[\/latex] by dividing the second term of the series by the first.\r\n<div style=\"text-align: center;\">[latex]r=\\frac{-4}{8}=-\\frac{1}{2}[\/latex]<\/div>\r\nSubstitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{11}&amp;=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\r\n<div style=\"text-align: center;\">[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]<\/div>\r\nWe can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is 6, so [latex]n=6[\/latex].\r\n\r\nSubstitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{6}&amp;=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nUse the formula to find the indicated partial sum of each geometric series.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots [\/latex]\r\n\r\n[reveal-answer q=\"540772\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"540772\"]\r\n\r\n[latex]\\approx 2,000.00[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]\r\n\r\n[reveal-answer q=\"915066\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"915066\"]\r\n\r\n9,840\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]29766[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172729[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Solving an Application Problem with a Geometric Series<\/h3>\r\nAt a new job, an employee\u2019s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.\r\n\r\n[reveal-answer q=\"474407\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"474407\"]\r\n\r\nThe problem can be represented by a geometric series with [latex]{a}_{1}=26,750[\/latex]; [latex]n=5[\/latex]; and [latex]r=1.016[\/latex]. Substitute values for [latex]{a}_{1}[\/latex], [latex]r[\/latex], and [latex]n[\/latex] into the formula and simplify to find the total amount earned at the end of 5 years.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&amp;=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{5}&amp;=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03 \\end{align}[\/latex]<\/p>\r\nHe will have earned a total of $138,099.03 by the end of 5 years.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nAt a new job, an employee\u2019s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?\r\n\r\n[reveal-answer q=\"442418\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"442418\"]\r\n\r\n$275,513.31\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Using the Formula for the Sum of an Infinite Geometric Series<\/h2>\r\nThus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first [latex]n[\/latex] terms. An <strong>infinite series<\/strong> is the sum of the terms of an infinite sequence. An example of an infinite series is [latex]2+4+6+8+..[\/latex].\r\n\r\nThis series can also be written in summation notation as [latex]\\sum _{k=1}^{\\infty }2k[\/latex], where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series <strong>diverges<\/strong>.\r\n<h2>Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h2>\r\nIf the terms of an <strong>infinite geometric series<\/strong> approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:\r\n<div style=\"text-align: center;\">[latex]1+0.2+0.04+0.008+0.0016+..[\/latex].<\/div>\r\nThe common ratio [latex]r\\text{ = 0}\\text{.2}[\/latex].\u00a0As [latex]n[\/latex] gets very large, the values of [latex]{r}^{n}[\/latex] get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with [latex]-1 < r < 1[\/latex] approach 0; the sum of a geometric series is defined when [latex]-1 < r < 1[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h3>\r\nThe sum of an infinite series is defined if the series is geometric and [latex]-1 < r < 1[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the first several terms of an infinite series, determine if the sum of the series exists.<\/h3>\r\n<ol>\r\n \t<li>Find the ratio of the second term to the first term.<\/li>\r\n \t<li>Find the ratio of the third term to the second term.<\/li>\r\n \t<li>Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.<\/li>\r\n \t<li>If a common ratio, [latex]r[\/latex], was found in step 3, check to see if [latex]-1 < r < 1[\/latex] . If so, the sum is defined. If not, the sum is not defined.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Determining Whether the Sum of an Infinite Series is Defined<\/h3>\r\nDetermine whether the sum of each infinite series is defined.\r\n<ol>\r\n \t<li>[latex]\\text{12 + 8 + 4 + }\\dots [\/latex]<\/li>\r\n \t<li>[latex]\\frac{3}{4}+\\frac{1}{2}+\\frac{1}{3}+..[\/latex].<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{\\infty }27\\cdot {\\left(\\frac{1}{3}\\right)}^{k}[\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{\\infty }5k[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"648214\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"648214\"]\r\n<ol>\r\n \t<li>The ratio of the second term to the first is [latex]\\frac{\\text{2}}{\\text{3}}[\/latex],\r\nwhich is not the same as the ratio of the third term to the second, [latex]\\frac{1}{2}[\/latex]. The series is not geometric.<\/li>\r\n \t<li>The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of [latex]\\frac{2}{3}\\text{.}[\/latex] The sum of the infinite series is defined.<\/li>\r\n \t<li>The given formula is exponential with a base of [latex]\\frac{1}{3}[\/latex]; the series is geometric with a common ratio of [latex]\\frac{1}{3}\\text{.}[\/latex] The sum of the infinite series is defined.<\/li>\r\n \t<li>The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nDetermine whether the sum of the infinite series is defined.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\frac{1}{3}+\\frac{1}{2}+\\frac{3}{4}+\\frac{9}{8}+..[\/latex].\r\n\r\n[reveal-answer q=\"296854\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296854\"]\r\n\r\nThe sum is not defined. It is not geometric.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]24+\\left(-12\\right)+6+\\left(-3\\right)+..[\/latex].\r\n\r\n[reveal-answer q=\"838480\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"838480\"]\r\n\r\nThe sum of the infinite series is defined.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\sum _{k=1}^{\\infty }15\\cdot {\\left(-0.3\\right)}^{k}[\/latex]\r\n\r\n[reveal-answer q=\"412528\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"412528\"]\r\n\r\nThe sum of the infinite series is defined.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Finding Sums of Infinite Series<\/h2>\r\nWhen the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first [latex]n[\/latex] terms of a geometric series.\r\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/div>\r\nWe will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]{r}^{n}[\/latex] as [latex]n[\/latex] increases?\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\left(\\frac{1}{2}\\right)}^{2}=\\frac{1}{4}\\\\ &amp;{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{1}{8}\\\\ &amp;{\\left(\\frac{1}{2}\\right)}^{4}=\\frac{1}{16}\\end{align}[\/latex]<\/div>\r\nThe value of [latex]{r}^{n}[\/latex] decreases rapidly. What happens for greater values of [latex]n?[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\begin{align}&amp;{\\left(\\frac{1}{2}\\right)}^{10}=\\frac{1}{1\\text{,}024} \\\\ &amp;{\\left(\\frac{1}{2}\\right)}^{20}=\\frac{1}{1\\text{,}048\\text{,}576} \\\\ &amp;{\\left(\\frac{1}{2}\\right)}^{30}=\\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/div>\r\nAs [latex]n[\/latex] gets very large, [latex]{r}^{n}[\/latex] gets very small. We say that, as [latex]n[\/latex] increases without bound, [latex]{r}^{n}[\/latex] approaches 0. As [latex]{r}^{n}[\/latex] approaches 0, [latex]1-{r}^{n}[\/latex] approaches 1. When this happens, the numerator approaches [latex]{a}_{1}[\/latex]. This give us a formula for the sum of an infinite geometric series.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Formula for the Sum of an Infinite Geometric Series<\/h3>\r\nThe formula for the sum of an infinite geometric series with [latex]-1&lt;r&lt;1[\/latex] is\r\n<p style=\"text-align: center;\">[latex]S=\\frac{{a}_{1}}{1-r}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an infinite geometric series, find its sum.<\/h3>\r\n<ol>\r\n \t<li>Identify [latex]{a}_{1}[\/latex] and [latex]r[\/latex].<\/li>\r\n \t<li>Confirm that [latex]-1&lt;r&lt;1[\/latex].<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1}[\/latex] and [latex]r[\/latex] into the formula, [latex]S=\\frac{{a}_{1}}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]S[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Finding the Sum of an Infinite Geometric Series<\/h3>\r\nFind the sum, if it exists, for the following:\r\n<ol>\r\n \t<li>[latex]10+9+8+7+\\dots [\/latex]<\/li>\r\n \t<li>[latex]248.6+99.44+39.776+\\text{ }\\dots [\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{\\infty }4\\text{,}374\\cdot {\\left(-\\frac{1}{3}\\right)}^{k - 1}[\/latex]<\/li>\r\n \t<li>[latex]\\sum _{k=1}^{\\infty }\\frac{1}{9}\\cdot {\\left(\\frac{4}{3}\\right)}^{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"197649\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"197649\"]\r\n<ol>\r\n \t<li>There is not a constant ratio; the series is not geometric.<\/li>\r\n \t<li>There is a constant ratio; the series is geometric. [latex]{a}_{1}=248.6[\/latex] and [latex]r=\\frac{99.44}{248.6}=0.4[\/latex], so the sum exists. Substitute [latex]{a}_{1}=248.6[\/latex] and [latex]r=0.4[\/latex] into the formula and simplify to find the sum:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}S&amp;=\\frac{{a}_{1}}{1-r} \\\\ S&amp;=\\frac{248.6}{1 - 0.4}=414.\\overline{3} \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>The formula is exponential, so the series is geometric with [latex]r=-\\frac{1}{3}[\/latex]. Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula:\r\n<div style=\"text-align: center;\">[latex]{a}_{1}=4\\text{,}374\\cdot {\\left(-\\frac{1}{3}\\right)}^{1 - 1}=4\\text{,}374[\/latex]<\/div>\r\nSubstitute [latex]{a}_{1}=4\\text{,}374[\/latex] and [latex]r=-\\frac{1}{3}[\/latex] into the formula, and simplify to find the sum:\r\n<div style=\"text-align: center;\">[latex]\\begin{align}S&amp;=\\frac{{a}_{1}}{1-r}\\\\ S&amp;=\\frac{4\\text{,}374}{1-\\left(-\\frac{1}{3}\\right)}=3\\text{,}280.5 \\end{align}[\/latex]<\/div><\/li>\r\n \t<li>The formula is exponential, so the series is geometric, but [latex]r&gt;1[\/latex]. The sum does not exist.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Finding an Equivalent Fraction for a Repeating Decimal<\/h3>\r\nFind an equivalent fraction for the repeating decimal [latex]0.\\overline{3}[\/latex]\r\n\r\n[reveal-answer q=\"280993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"280993\"]\r\n\r\nWe notice the repeating decimal [latex]0.\\overline{3}=0.333..[\/latex]. so we can rewrite the repeating decimal as a sum of terms.\r\n<p style=\"text-align: center;\">[latex]0.\\overline{3}=0.3+0.03+0.003+..[\/latex].<\/p>\r\nLooking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/09\/Screen-Shot-2015-09-15-at-10.22.06-PM.png\"><img class=\"aligncenter wp-image-11435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183629\/Screen-Shot-2015-09-15-at-10.22.06-PM.png\" alt=\"0.3 repeating equals 0.3 + 0.1 + 0.3 (the first term) + 0.1. (0.1)(0.3) (the second term)\" width=\"403\" height=\"86\" \/><\/a>\r\n\r\nNotice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have\r\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}=\\frac{0.3}{1 - 0.1}=\\frac{0.3}{0.9}=\\frac{1}{3}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nFind the sum, if it exists.\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]2+\\frac{2}{3}+\\frac{2}{9}+..[\/latex].\r\n\r\n[reveal-answer q=\"903460\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"903460\"]\r\n\r\n3\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\sum _{k=1}^{\\infty }0.76k+1[\/latex]\r\n\r\n[reveal-answer q=\"814253\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"814253\"]\r\n\r\nThe series is not geometric.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[latex]\\sum _{k=1}^{\\infty }{\\left(-\\frac{3}{8}\\right)}^{k}[\/latex]\r\n\r\n[reveal-answer q=\"462522\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"462522\"]\r\n\r\n[latex]-\\frac{3}{11}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]172852[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Annuity Problems<\/h2>\r\nAt the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An <strong>annuity<\/strong> is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% <strong>annual interest<\/strong>, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.\r\n\r\nWe can find the value of the annuity right after the last deposit by using a geometric series with [latex]{a}_{1}=50[\/latex] and [latex]r=100.5%=1.005[\/latex]. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.\r\n\r\nWe can find the value of the annuity after [latex]n[\/latex] deposits using the formula for the sum of the first [latex]n[\/latex] terms of a geometric series. In 6 years, there are 72 months, so [latex]n=72[\/latex]. We can substitute [latex]{a}_{1}=50, r=1.005, \\text{and} n=72[\/latex] into the formula, and simplify to find the value of the annuity after 6 years.\r\n<div style=\"text-align: center;\">[latex]{S}_{72}=\\frac{50\\left(1-{1.005}^{72}\\right)}{1 - 1.005}\\approx 4\\text{,}320.44[\/latex]<\/div>\r\nAfter the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of [latex]72\\left(50\\right) = $3,600[\/latex]. This means that because of the annuity, the couple earned $720.44 interest in their college fund.\r\n<div class=\"textbox\">\r\n<h3>How To: Given an initial deposit and an interest rate, find the value of an annuity.<\/h3>\r\n<ol>\r\n \t<li>Determine [latex]{a}_{1}[\/latex], the value of the initial deposit.<\/li>\r\n \t<li>Determine [latex]n[\/latex], the number of deposits.<\/li>\r\n \t<li>Determine [latex]r[\/latex].\r\n<ol>\r\n \t<li>Divide the annual interest rate by the number of times per year that interest is compounded.<\/li>\r\n \t<li>Add 1 to this amount to find [latex]r[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Substitute values for [latex]{a}_{1}\\text{,}r,\\text{ and }n[\/latex]\r\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\r\n \t<li>Simplify to find [latex]{S}_{n}[\/latex], the value of the annuity after [latex]n[\/latex] deposits.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Solving an Annuity Problem<\/h3>\r\nA deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?\r\n\r\n[reveal-answer q=\"385976\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"385976\"]\r\n\r\nThe value of the initial deposit is $100, so [latex]{a}_{1}=100[\/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[\/latex]. To find [latex]r[\/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.\r\n<div style=\"text-align: center;\">[latex]r=1+\\frac{0.09}{12}=1.0075[\/latex]<\/div>\r\nSubstitute [latex]{a}_{1}=100\\text{,}r=1.0075\\text{,}\\text{and}n=120[\/latex] into the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, and simplify to find the value of the annuity.\r\n<div style=\"text-align: center;\">[latex]{S}_{120}=\\frac{100\\left(1-{1.0075}^{120}\\right)}{1 - 1.0075}\\approx 19\\text{,}351.43[\/latex]<\/div>\r\nSo the account has $19,351.43 after the last deposit is made.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nAt the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?\r\n\r\n[reveal-answer q=\"753036\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"753036\"]\r\n\r\n$92,408.18\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Key Equations<\/h2>\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>sum of the first [latex]n[\/latex]\r\nterms of an arithmetic series<\/td>\r\n<td>[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>sum of the first [latex]n[\/latex]\r\nterms of a geometric series<\/td>\r\n<td>[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>sum of an infinite geometric series with [latex]-1&lt;r&lt;\\text{ }1[\/latex]<\/td>\r\n<td>[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>The sum of the terms in a sequence is called a series.<\/li>\r\n \t<li>A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum.<\/li>\r\n \t<li>The sum of the terms in an arithmetic sequence is called an arithmetic series.<\/li>\r\n \t<li>The sum of the first [latex]n[\/latex] terms of an arithmetic series can be found using a formula.<\/li>\r\n \t<li>The sum of the terms in a geometric sequence is called a geometric series.<\/li>\r\n \t<li>The sum of the first [latex]n[\/latex] terms of a geometric series can be found using a formula.<\/li>\r\n \t<li>The sum of an infinite series exists if the series is geometric with [latex]-1&lt;r&lt;1[\/latex].<\/li>\r\n \t<li>If the sum of an infinite series exists, it can be found using a formula.<\/li>\r\n \t<li>An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137726792\" class=\"definition\">\r\n \t<dt>annuity<\/dt>\r\n \t<dd id=\"fs-id1165137726797\">an investment in which the purchaser makes a sequence of periodic, equal payments<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137726801\" class=\"definition\">\r\n \t<dt>arithmetic series<\/dt>\r\n \t<dd id=\"fs-id1165137726806\">the sum of the terms in an arithmetic sequence<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137726811\" class=\"definition\">\r\n \t<dt>diverge<\/dt>\r\n \t<dd id=\"fs-id1165134031382\">a series is said to diverge if the sum is not a real number<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134031386\" class=\"definition\">\r\n \t<dt>geometric series<\/dt>\r\n \t<dd id=\"fs-id1165134031391\">the sum of the terms in a geometric sequence<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134031396\" class=\"definition\">\r\n \t<dt>index of summation<\/dt>\r\n \t<dd id=\"fs-id1165134031401\">in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137737862\" class=\"definition\">\r\n \t<dt>infinite series<\/dt>\r\n \t<dd id=\"fs-id1165137737867\">the sum of the terms in an infinite sequence<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137737872\" class=\"definition\">\r\n \t<dt>lower limit of summation<\/dt>\r\n \t<dd id=\"fs-id1165137737877\">the number used in the explicit formula to find the first term in a series<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137737881\" class=\"definition\">\r\n \t<dt>nth partial sum<\/dt>\r\n \t<dd id=\"fs-id1165135471113\">the sum of the first [latex]n[\/latex] terms of a sequence<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135471124\" class=\"definition\">\r\n \t<dt>series<\/dt>\r\n \t<dd id=\"fs-id1165135471129\">the sum of the terms in a sequence<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135471133\" class=\"definition\">\r\n \t<dt>summation notation<\/dt>\r\n \t<dd id=\"fs-id1165135471138\">a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137762744\" class=\"definition\">\r\n \t<dt>upper limit of summation<\/dt>\r\n \t<dd id=\"fs-id1165137762749\">the number used in the explicit formula to find the last term in a series<\/dd>\r\n<\/dl>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li style=\"font-weight: 400;\">Use summation notation.<\/li>\n<li style=\"font-weight: 400;\">Use the formula for the sum of the \ufb01rst n terms of an arithmetic series.<\/li>\n<li style=\"font-weight: 400;\">Use the formula for the sum of the \ufb01rst n terms of a geometric series.<\/li>\n<li style=\"font-weight: 400;\">Use the formula for the sum of an in\ufb01nite geometric series.<\/li>\n<li style=\"font-weight: 400;\">Solve word problems involving series.<\/li>\n<\/ul>\n<\/div>\n<h2>Using Summation Notation<\/h2>\n<p>To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a <strong>series<\/strong>. Consider, for example, the following series.<\/p>\n<div style=\"text-align: center;\">[latex]3+7+11+15+19+\\dots[\/latex]<\/div>\n<p>The <strong> [latex]n\\text{th }[\/latex] partial sum<\/strong> of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation [latex]\\text{ }{S}_{n}\\text{ }[\/latex] represents the partial sum.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{S}_{1}=3\\\\ &{S}_{2}=3+7=10\\\\ &{S}_{3}=3+7+11=21\\\\ &{S}_{4}=3+7+11+15=36\\end{align}[\/latex]<\/div>\n<p><strong>Summation notation <\/strong>is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter <strong>sigma<\/strong>, [latex]\\sigma[\/latex], to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the <strong>index of summation <\/strong>is written below the sigma. The index of summation is set equal to the <strong>lower limit of summation<\/strong>, which is the number used to generate the first term in the series. The number above the sigma, called the <strong>upper limit of summation<\/strong>, is the number used to generate the last term in a series.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183627\/CNX_Precalc_Figure_11_04_001n2.jpg\" alt=\"Explanation of summation notion as described in the text.\" \/><\/p>\n<p>If we interpret the given notation, we see that it asks us to find the sum of the terms in the series [latex]{a}_{k}=2k[\/latex] for [latex]k=1[\/latex] through [latex]k=5[\/latex]. We can begin by substituting the terms for [latex]k[\/latex] and listing out the terms of this series.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l} {a}_{1}=2\\left(1\\right)=2 \\\\ {a}_{2}=2\\left(2\\right)=4\\hfill \\\\ {a}_{3}=2\\left(3\\right)=6\\hfill \\\\ {a}_{4}=2\\left(4\\right)=8\\hfill \\\\ {a}_{5}=2\\left(5\\right)=10\\hfill \\end{array}[\/latex]<\/div>\n<p>We can find the sum of the series by adding the terms:<\/p>\n<div style=\"text-align: center;\">[latex]\\sum _{k=1}^{5}2k=2+4+6+8+10=30[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Summation Notation<\/h3>\n<p>The sum of the first [latex]n[\/latex] terms of a <strong>series <\/strong>can be expressed in <strong>summation notation<\/strong> as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\sum _{k=1}^{n}{a}_{k}[\/latex]<\/p>\n<p>This notation tells us to find the sum of [latex]{a}_{k}[\/latex] from [latex]k=1[\/latex] to [latex]k=n[\/latex].<\/p>\n<p>[latex]k[\/latex] is called the <strong>index of summation<\/strong>, 1 is the <strong>lower limit of summation<\/strong>, and [latex]n[\/latex] is the <strong>upper limit of summation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h3>Does the lower limit of summation have to be 1?<\/h3>\n<p><em>No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given summation notation for a series, evaluate the value.<\/h3>\n<ol>\n<li>Identify the lower limit of summation.<\/li>\n<li>Identify the upper limit of summation.<\/li>\n<li>Substitute each value of [latex]k[\/latex] from the lower limit to the upper limit into the formula.<\/li>\n<li>Add to find the sum.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 1: Using Summation Notation<\/h3>\n<p>Evaluate [latex]\\sum _{k=3}^{7}{k}^{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14937\">Show Solution<\/span><\/p>\n<div id=\"q14937\" class=\"hidden-answer\" style=\"display: none\">\n<p>According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of [latex]{k}^{2}[\/latex] from [latex]k=3[\/latex] to [latex]k=7[\/latex]. We find the terms of the series by substituting [latex]k=3\\text{,}4\\text{,}5\\text{,}6[\/latex], and [latex]7[\/latex] into the function [latex]{k}^{2}[\/latex]. We add the terms to find the sum.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sum _{k=3}^{7}{k}^{2}& ={3}^{2}+{4}^{2}+{5}^{2}+{6}^{2}+{7}^{2}\\hfill \\\\ \\hfill & =9+16+25+36+49\\hfill \\\\ \\hfill & =135\\hfill \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]\\sum _{k=2}^{5}\\left(3k - 1\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q812548\">Show Solution<\/span><\/p>\n<div id=\"q812548\" class=\"hidden-answer\" style=\"display: none\">\n<p>38<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172726\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172726&theme=oea&iframe_resize_id=ohm172726\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Formula for Arithmetic Series<\/h2>\n<p>Just as we studied special types of sequences, we will look at special types of series. Recall that an <strong>arithmetic sequence<\/strong> is a sequence in which the difference between any two consecutive terms is the <strong>common difference<\/strong>, [latex]d[\/latex]. The sum of the terms of an arithmetic sequence is called an <strong>arithmetic series<\/strong>. We can write the sum of the first [latex]n[\/latex] terms of an arithmetic series as:<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}[\/latex].<\/div>\n<p>We can also reverse the order of the terms and write the sum as<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}[\/latex].<\/div>\n<p>If we add these two expressions for the sum of the first [latex]n[\/latex] terms of an arithmetic series, we can derive a formula for the sum of the first [latex]n[\/latex] terms of any arithmetic series.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&={a}_{1}+\\left({a}_{1}+d\\right)+\\left({a}_{1}+2d\\right)+...+\\left({a}_{n}-d\\right)+{a}_{n}\\hfill \\\\ +{S}_{n}&={a}_{n}+\\left({a}_{n}-d\\right)+\\left({a}_{n}-2d\\right)+...+\\left({a}_{1}+d\\right)+{a}_{1}\\\\ \\hline 2{S}_{n}&=\\left({a}_{1}+{a}_{n}\\right)+\\left({a}_{1}+{a}_{n}\\right)+...+\\left({a}_{1}+{a}_{n}\\right)\\end{align}[\/latex]<\/div>\n<p>Because there are [latex]n[\/latex] terms in the series, we can simplify this sum to<\/p>\n<div style=\"text-align: center;\">[latex]2{S}_{n}=n\\left({a}_{1}+{a}_{n}\\right)[\/latex].<\/div>\n<p>We divide by 2 to find the formula for the sum of the first [latex]n[\/latex] terms of an arithmetic series.<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of an Arithmetic Series<\/h3>\n<p>An <strong>arithmetic series<\/strong> is the sum of the terms of an arithmetic sequence. The formula for the sum of the first [latex]n[\/latex] terms of an arithmetic sequence is<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given terms of an arithmetic series, find the sum of the first [latex]n[\/latex] terms.<\/h3>\n<ol>\n<li>Identify [latex]{a}_{1}[\/latex] and [latex]{a}_{n}[\/latex].<\/li>\n<li>Determine [latex]n[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1}\\text{, }{a}_{n}[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Finding the First <em>n<\/em> Terms of an Arithmetic Series<\/h3>\n<p>Find the sum of each arithmetic series.<\/p>\n<ol>\n<li>[latex]5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32[\/latex]<\/li>\n<li>[latex]20 + 15 + 10 +\\dots + -50[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{12}3k - 8[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535733\">Show Solution<\/span><\/p>\n<div id=\"q535733\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>We are given [latex]{a}_{1}=5[\/latex] and [latex]{a}_{n}=32[\/latex].Count the number of terms in the sequence to find [latex]n=10[\/latex]. Substitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{10}&=\\frac{10\\left(5+32\\right)}{2}=185 \\end{align}[\/latex]<\/div>\n<\/li>\n<li>We are given [latex]{a}_{1}=20[\/latex] and [latex]{a}_{n}=-50[\/latex].Use the formula for the general term of an arithmetic sequence to find [latex]n[\/latex].\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{n}&={a}_{1}+\\left(n - 1\\right)d \\\\ -50&=20+\\left(n - 1\\right)\\left(-5\\right)\\\\ -70&=\\left(n - 1\\right)\\left(-5\\right) \\\\ 14&=n - 1\\\\ 15&=n\\end{align}[\/latex]<\/div>\n<p>Substitute values for [latex]{a}_{1},{a}_{n}\\text{,}n[\/latex] into the formula and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align} {S}_{n}&=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{15}&=\\frac{15\\left(20 - 50\\right)}{2}=-225\\end{align}[\/latex]<\/div>\n<\/li>\n<li>To find [latex]{a}_{1}[\/latex], substitute [latex]k=1[\/latex] into the given explicit formula.\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{k}&=3k - 8 \\\\ {a}_{1}&=3\\left(1\\right)-8=-5\\end{align}[\/latex]<\/div>\n<p>We are given that [latex]n=12[\/latex]. To find [latex]{a}_{12}[\/latex], substitute [latex]k=12[\/latex] into the given explicit formula.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{a}_{k}&=3k - 8 \\\\ {a}_{12}&=3\\left(12\\right)-8=28 \\end{align}[\/latex]<\/div>\n<p>Substitute values for [latex]{a}_{1},{a}_{n}[\/latex], and [latex]n[\/latex] into the formula and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{12}&=\\frac{12\\left(-5+28\\right)}{2}=138 \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Use the formula to find the sum of each arithmetic series.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\text{1}\\text{.4 + 1}\\text{.6 + 1}\\text{.8 + 2}\\text{.0 + 2}\\text{.2 + 2}\\text{.4 + 2}\\text{.6 + 2}\\text{.8 + 3}\\text{.0 + 3}\\text{.2 + 3}\\text{.4}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q715922\">Show Solution<\/span><\/p>\n<div id=\"q715922\" class=\"hidden-answer\" style=\"display: none\">\n<p>26.4<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\text{13 + 21 + 29 + }\\dots \\text{+ 69}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q81811\">Show Solution<\/span><\/p>\n<div id=\"q81811\" class=\"hidden-answer\" style=\"display: none\">\n<p>328<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\sum _{k=1}^{10}5 - 6k[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16304\">Show Solution<\/span><\/p>\n<div id=\"q16304\" class=\"hidden-answer\" style=\"display: none\">\n<p>-280<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm23739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=23739&theme=oea&iframe_resize_id=ohm23739\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Solving Application Problems with Arithmetic Series<\/h3>\n<p>On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924199\">Show Solution<\/span><\/p>\n<div id=\"q924199\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem can be modeled by an arithmetic series with [latex]{a}_{1}=\\frac{1}{2}[\/latex] and [latex]d=\\frac{1}{4}[\/latex]. We are looking for the total number of miles walked after 8 weeks, so we know that [latex]n=8[\/latex], and we are looking for [latex]{S}_{8}[\/latex]. To find [latex]{a}_{8}[\/latex], we can use the explicit formula for an arithmetic sequence.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {a}_{n}&={a}_{1}+d\\left(n - 1\\right) \\\\ {a}_{8}&=\\frac{1}{2}+\\frac{1}{4}\\left(8 - 1\\right)=\\frac{9}{4} \\end{align}[\/latex]<\/p>\n<p>We can now use the formula for arithmetic series.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} {S}_{n}&=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2} \\\\ {S}_{8}&=\\frac{8\\left(\\frac{1}{2}+\\frac{9}{4}\\right)}{2}=11 \\end{align}[\/latex]<\/p>\n<p>She will have walked a total of 11 miles.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. After 12 weeks, how much has he earned?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q505026\">Show Solution<\/span><\/p>\n<div id=\"q505026\" class=\"hidden-answer\" style=\"display: none\">\n<p>$2,025<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using the Formula for Geometric Series<\/h2>\n<p>Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a <strong>geometric series<\/strong>. Recall that a <strong>geometric sequence<\/strong> is a sequence in which the ratio of any two consecutive terms is the <strong>common ratio<\/strong>, [latex]r[\/latex]. We can write the sum of the first [latex]n[\/latex] terms of a geometric series as<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[\/latex].<\/div>\n<p>Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first [latex]n[\/latex] terms of a geometric series. We will begin by multiplying both sides of the equation by [latex]r[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]r{S}_{n}=r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}[\/latex]<\/div>\n<p>Next, we subtract this equation from the original equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1} \\\\ -r{S}_{n}&=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right) \\\\ \\hline\\left(1-r\\right){S}_{n}&={a}_{1}-{r}^{n}{a}_{1}\\end{align}[\/latex]<\/div>\n<p>Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for [latex]{S}_{n}[\/latex], divide both sides by [latex]\\left(1-r\\right)[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{ r}\\ne \\text{1}[\/latex]<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Sum of the First <em>n<\/em> Terms of a Geometric Series<\/h3>\n<p>A <strong>geometric series<\/strong> is the sum of the terms in a geometric sequence. The formula for the sum of the first [latex]n[\/latex] terms of a geometric sequence is represented as<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\text{, r}\\ne \\text{1}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a geometric series, find the sum of the first <em>n<\/em> terms.<\/h3>\n<ol>\n<li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Finding the First <em>n<\/em> Terms of a Geometric Series<\/h3>\n<p>Use the formula to find the indicated partial sum of each geometric series.<\/p>\n<ol>\n<li>[latex]{S}_{11}[\/latex] for the series [latex]\\text{ 8 + -4 + 2 + }\\dots[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{6}3\\cdot {2}^{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q792584\">Show Solution<\/span><\/p>\n<div id=\"q792584\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]{a}_{1}=8[\/latex], and we are given that [latex]n=11[\/latex].We can find [latex]r[\/latex] by dividing the second term of the series by the first.\n<div style=\"text-align: center;\">[latex]r=\\frac{-4}{8}=-\\frac{1}{2}[\/latex]<\/div>\n<p>Substitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{11}&=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336 \\end{align}[\/latex]<\/div>\n<\/li>\n<li>Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula.\n<div style=\"text-align: center;\">[latex]{a}_{1}=3\\cdot {2}^{1}=6[\/latex]<\/div>\n<p>We can see from the given explicit formula that [latex]r=2[\/latex]. The upper limit of summation is 6, so [latex]n=6[\/latex].<\/p>\n<p>Substitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{6}&=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378 \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Use the formula to find the indicated partial sum of each geometric series.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q540772\">Show Solution<\/span><\/p>\n<div id=\"q540772\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\approx 2,000.00[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q915066\">Show Solution<\/span><\/p>\n<div id=\"q915066\" class=\"hidden-answer\" style=\"display: none\">\n<p>9,840<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm29766\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=29766&theme=oea&iframe_resize_id=ohm29766\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172729\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172729&theme=oea&iframe_resize_id=ohm172729\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Solving an Application Problem with a Geometric Series<\/h3>\n<p>At a new job, an employee\u2019s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q474407\">Show Solution<\/span><\/p>\n<div id=\"q474407\" class=\"hidden-answer\" style=\"display: none\">\n<p>The problem can be represented by a geometric series with [latex]{a}_{1}=26,750[\/latex]; [latex]n=5[\/latex]; and [latex]r=1.016[\/latex]. Substitute values for [latex]{a}_{1}[\/latex], [latex]r[\/latex], and [latex]n[\/latex] into the formula and simplify to find the total amount earned at the end of 5 years.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{S}_{n}&=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r} \\\\ {S}_{5}&=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03 \\end{align}[\/latex]<\/p>\n<p>He will have earned a total of $138,099.03 by the end of 5 years.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>At a new job, an employee\u2019s starting salary is $32,100. She receives a 2% annual raise. How much will she have earned by the end of 8 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q442418\">Show Solution<\/span><\/p>\n<div id=\"q442418\" class=\"hidden-answer\" style=\"display: none\">\n<p>$275,513.31<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Using the Formula for the Sum of an Infinite Geometric Series<\/h2>\n<p>Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first [latex]n[\/latex] terms. An <strong>infinite series<\/strong> is the sum of the terms of an infinite sequence. An example of an infinite series is [latex]2+4+6+8+..[\/latex].<\/p>\n<p>This series can also be written in summation notation as [latex]\\sum _{k=1}^{\\infty }2k[\/latex], where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series <strong>diverges<\/strong>.<\/p>\n<h2>Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h2>\n<p>If the terms of an <strong>infinite geometric series<\/strong> approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0:<\/p>\n<div style=\"text-align: center;\">[latex]1+0.2+0.04+0.008+0.0016+..[\/latex].<\/div>\n<p>The common ratio [latex]r\\text{ = 0}\\text{.2}[\/latex].\u00a0As [latex]n[\/latex] gets very large, the values of [latex]{r}^{n}[\/latex] get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with [latex]-1 < r < 1[\/latex] approach 0; the sum of a geometric series is defined when [latex]-1 < r < 1[\/latex].\n\n\n<div class=\"textbox\">\n<h3>A General Note: Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h3>\n<p>The sum of an infinite series is defined if the series is geometric and [latex]-1 < r < 1[\/latex].\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the first several terms of an infinite series, determine if the sum of the series exists.<\/h3>\n<ol>\n<li>Find the ratio of the second term to the first term.<\/li>\n<li>Find the ratio of the third term to the second term.<\/li>\n<li>Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric.<\/li>\n<li>If a common ratio, [latex]r[\/latex], was found in step 3, check to see if [latex]-1 < r < 1[\/latex] . If so, the sum is defined. If not, the sum is not defined.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Determining Whether the Sum of an Infinite Series is Defined<\/h3>\n<p>Determine whether the sum of each infinite series is defined.<\/p>\n<ol>\n<li>[latex]\\text{12 + 8 + 4 + }\\dots[\/latex]<\/li>\n<li>[latex]\\frac{3}{4}+\\frac{1}{2}+\\frac{1}{3}+..[\/latex].<\/li>\n<li>[latex]\\sum _{k=1}^{\\infty }27\\cdot {\\left(\\frac{1}{3}\\right)}^{k}[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{\\infty }5k[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q648214\">Show Solution<\/span><\/p>\n<div id=\"q648214\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>The ratio of the second term to the first is [latex]\\frac{\\text{2}}{\\text{3}}[\/latex],<br \/>\nwhich is not the same as the ratio of the third term to the second, [latex]\\frac{1}{2}[\/latex]. The series is not geometric.<\/li>\n<li>The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of [latex]\\frac{2}{3}\\text{.}[\/latex] The sum of the infinite series is defined.<\/li>\n<li>The given formula is exponential with a base of [latex]\\frac{1}{3}[\/latex]; the series is geometric with a common ratio of [latex]\\frac{1}{3}\\text{.}[\/latex] The sum of the infinite series is defined.<\/li>\n<li>The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Determine whether the sum of the infinite series is defined.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\frac{1}{3}+\\frac{1}{2}+\\frac{3}{4}+\\frac{9}{8}+..[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q296854\">Show Solution<\/span><\/p>\n<div id=\"q296854\" class=\"hidden-answer\" style=\"display: none\">\n<p>The sum is not defined. It is not geometric.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]24+\\left(-12\\right)+6+\\left(-3\\right)+..[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q838480\">Show Solution<\/span><\/p>\n<div id=\"q838480\" class=\"hidden-answer\" style=\"display: none\">\n<p>The sum of the infinite series is defined.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\sum _{k=1}^{\\infty }15\\cdot {\\left(-0.3\\right)}^{k}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q412528\">Show Solution<\/span><\/p>\n<div id=\"q412528\" class=\"hidden-answer\" style=\"display: none\">\n<p>The sum of the infinite series is defined.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Finding Sums of Infinite Series<\/h2>\n<p>When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first [latex]n[\/latex] terms of a geometric series.<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex]<\/div>\n<p>We will examine an infinite series with [latex]r=\\frac{1}{2}[\/latex]. What happens to [latex]{r}^{n}[\/latex] as [latex]n[\/latex] increases?<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{\\left(\\frac{1}{2}\\right)}^{2}=\\frac{1}{4}\\\\ &{\\left(\\frac{1}{2}\\right)}^{3}=\\frac{1}{8}\\\\ &{\\left(\\frac{1}{2}\\right)}^{4}=\\frac{1}{16}\\end{align}[\/latex]<\/div>\n<p>The value of [latex]{r}^{n}[\/latex] decreases rapidly. What happens for greater values of [latex]n?[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}&{\\left(\\frac{1}{2}\\right)}^{10}=\\frac{1}{1\\text{,}024} \\\\ &{\\left(\\frac{1}{2}\\right)}^{20}=\\frac{1}{1\\text{,}048\\text{,}576} \\\\ &{\\left(\\frac{1}{2}\\right)}^{30}=\\frac{1}{1\\text{,}073\\text{,}741\\text{,}824} \\end{align}[\/latex]<\/div>\n<p>As [latex]n[\/latex] gets very large, [latex]{r}^{n}[\/latex] gets very small. We say that, as [latex]n[\/latex] increases without bound, [latex]{r}^{n}[\/latex] approaches 0. As [latex]{r}^{n}[\/latex] approaches 0, [latex]1-{r}^{n}[\/latex] approaches 1. When this happens, the numerator approaches [latex]{a}_{1}[\/latex]. This give us a formula for the sum of an infinite geometric series.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Formula for the Sum of an Infinite Geometric Series<\/h3>\n<p>The formula for the sum of an infinite geometric series with [latex]-1<r<1[\/latex] is\n\n\n<p style=\"text-align: center;\">[latex]S=\\frac{{a}_{1}}{1-r}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an infinite geometric series, find its sum.<\/h3>\n<ol>\n<li>Identify [latex]{a}_{1}[\/latex] and [latex]r[\/latex].<\/li>\n<li>Confirm that [latex]-1<r<1[\/latex].<\/li>\n<li>Substitute values for [latex]{a}_{1}[\/latex] and [latex]r[\/latex] into the formula, [latex]S=\\frac{{a}_{1}}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]S[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Finding the Sum of an Infinite Geometric Series<\/h3>\n<p>Find the sum, if it exists, for the following:<\/p>\n<ol>\n<li>[latex]10+9+8+7+\\dots[\/latex]<\/li>\n<li>[latex]248.6+99.44+39.776+\\text{ }\\dots[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{\\infty }4\\text{,}374\\cdot {\\left(-\\frac{1}{3}\\right)}^{k - 1}[\/latex]<\/li>\n<li>[latex]\\sum _{k=1}^{\\infty }\\frac{1}{9}\\cdot {\\left(\\frac{4}{3}\\right)}^{k}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q197649\">Show Solution<\/span><\/p>\n<div id=\"q197649\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>There is not a constant ratio; the series is not geometric.<\/li>\n<li>There is a constant ratio; the series is geometric. [latex]{a}_{1}=248.6[\/latex] and [latex]r=\\frac{99.44}{248.6}=0.4[\/latex], so the sum exists. Substitute [latex]{a}_{1}=248.6[\/latex] and [latex]r=0.4[\/latex] into the formula and simplify to find the sum:\n<div style=\"text-align: center;\">[latex]\\begin{align}S&=\\frac{{a}_{1}}{1-r} \\\\ S&=\\frac{248.6}{1 - 0.4}=414.\\overline{3} \\end{align}[\/latex]<\/div>\n<\/li>\n<li>The formula is exponential, so the series is geometric with [latex]r=-\\frac{1}{3}[\/latex]. Find [latex]{a}_{1}[\/latex] by substituting [latex]k=1[\/latex] into the given explicit formula:\n<div style=\"text-align: center;\">[latex]{a}_{1}=4\\text{,}374\\cdot {\\left(-\\frac{1}{3}\\right)}^{1 - 1}=4\\text{,}374[\/latex]<\/div>\n<p>Substitute [latex]{a}_{1}=4\\text{,}374[\/latex] and [latex]r=-\\frac{1}{3}[\/latex] into the formula, and simplify to find the sum:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}S&=\\frac{{a}_{1}}{1-r}\\\\ S&=\\frac{4\\text{,}374}{1-\\left(-\\frac{1}{3}\\right)}=3\\text{,}280.5 \\end{align}[\/latex]<\/div>\n<\/li>\n<li>The formula is exponential, so the series is geometric, but [latex]r>1[\/latex]. The sum does not exist.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Finding an Equivalent Fraction for a Repeating Decimal<\/h3>\n<p>Find an equivalent fraction for the repeating decimal [latex]0.\\overline{3}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q280993\">Show Solution<\/span><\/p>\n<div id=\"q280993\" class=\"hidden-answer\" style=\"display: none\">\n<p>We notice the repeating decimal [latex]0.\\overline{3}=0.333..[\/latex]. so we can rewrite the repeating decimal as a sum of terms.<\/p>\n<p style=\"text-align: center;\">[latex]0.\\overline{3}=0.3+0.03+0.003+..[\/latex].<\/p>\n<p>Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term.<\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/wp-content\/uploads\/sites\/145\/2015\/09\/Screen-Shot-2015-09-15-at-10.22.06-PM.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-11435 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27183629\/Screen-Shot-2015-09-15-at-10.22.06-PM.png\" alt=\"0.3 repeating equals 0.3 + 0.1 + 0.3 (the first term) + 0.1. (0.1)(0.3) (the second term)\" width=\"403\" height=\"86\" \/><\/a><\/p>\n<p>Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have<\/p>\n<p style=\"text-align: center;\">[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}=\\frac{0.3}{1 - 0.1}=\\frac{0.3}{0.9}=\\frac{1}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Find the sum, if it exists.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]2+\\frac{2}{3}+\\frac{2}{9}+..[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q903460\">Show Solution<\/span><\/p>\n<div id=\"q903460\" class=\"hidden-answer\" style=\"display: none\">\n<p>3<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\sum _{k=1}^{\\infty }0.76k+1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q814253\">Show Solution<\/span><\/p>\n<div id=\"q814253\" class=\"hidden-answer\" style=\"display: none\">\n<p>The series is not geometric.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>[latex]\\sum _{k=1}^{\\infty }{\\left(-\\frac{3}{8}\\right)}^{k}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q462522\">Show Solution<\/span><\/p>\n<div id=\"q462522\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\frac{3}{11}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm172852\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=172852&theme=oea&iframe_resize_id=ohm172852\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Annuity Problems<\/h2>\n<p>At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An <strong>annuity<\/strong> is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% <strong>annual interest<\/strong>, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added.<\/p>\n<p>We can find the value of the annuity right after the last deposit by using a geometric series with [latex]{a}_{1}=50[\/latex] and [latex]r=100.5%=1.005[\/latex]. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned.<\/p>\n<p>We can find the value of the annuity after [latex]n[\/latex] deposits using the formula for the sum of the first [latex]n[\/latex] terms of a geometric series. In 6 years, there are 72 months, so [latex]n=72[\/latex]. We can substitute [latex]{a}_{1}=50, r=1.005, \\text{and} n=72[\/latex] into the formula, and simplify to find the value of the annuity after 6 years.<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{72}=\\frac{50\\left(1-{1.005}^{72}\\right)}{1 - 1.005}\\approx 4\\text{,}320.44[\/latex]<\/div>\n<p>After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of [latex]72\\left(50\\right) = $3,600[\/latex]. This means that because of the annuity, the couple earned $720.44 interest in their college fund.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given an initial deposit and an interest rate, find the value of an annuity.<\/h3>\n<ol>\n<li>Determine [latex]{a}_{1}[\/latex], the value of the initial deposit.<\/li>\n<li>Determine [latex]n[\/latex], the number of deposits.<\/li>\n<li>Determine [latex]r[\/latex].\n<ol>\n<li>Divide the annual interest rate by the number of times per year that interest is compounded.<\/li>\n<li>Add 1 to this amount to find [latex]r[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>Substitute values for [latex]{a}_{1}\\text{,}r,\\text{ and }n[\/latex]<br \/>\ninto the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, [latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}[\/latex].<\/li>\n<li>Simplify to find [latex]{S}_{n}[\/latex], the value of the annuity after [latex]n[\/latex] deposits.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Solving an Annuity Problem<\/h3>\n<p>A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q385976\">Show Solution<\/span><\/p>\n<div id=\"q385976\" class=\"hidden-answer\" style=\"display: none\">\n<p>The value of the initial deposit is $100, so [latex]{a}_{1}=100[\/latex]. A total of 120 monthly deposits are made in the 10 years, so [latex]n=120[\/latex]. To find [latex]r[\/latex], divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit.<\/p>\n<div style=\"text-align: center;\">[latex]r=1+\\frac{0.09}{12}=1.0075[\/latex]<\/div>\n<p>Substitute [latex]{a}_{1}=100\\text{,}r=1.0075\\text{,}\\text{and}n=120[\/latex] into the formula for the sum of the first [latex]n[\/latex] terms of a geometric series, and simplify to find the value of the annuity.<\/p>\n<div style=\"text-align: center;\">[latex]{S}_{120}=\\frac{100\\left(1-{1.0075}^{120}\\right)}{1 - 1.0075}\\approx 19\\text{,}351.43[\/latex]<\/div>\n<p>So the account has $19,351.43 after the last deposit is made.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q753036\">Show Solution<\/span><\/p>\n<div id=\"q753036\" class=\"hidden-answer\" style=\"display: none\">\n<p>$92,408.18<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Equations<\/h2>\n<table>\n<tbody>\n<tr>\n<td>sum of the first [latex]n[\/latex]<br \/>\nterms of an arithmetic series<\/td>\n<td>[latex]{S}_{n}=\\frac{n\\left({a}_{1}+{a}_{n}\\right)}{2}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>sum of the first [latex]n[\/latex]<br \/>\nterms of a geometric series<\/td>\n<td>[latex]{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>sum of an infinite geometric series with [latex]-1<r<\\text{ }1[\/latex]<\/td>\n<td>[latex]{S}_{n}=\\frac{{a}_{1}}{1-r}\\cdot r\\ne 1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>The sum of the terms in a sequence is called a series.<\/li>\n<li>A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum.<\/li>\n<li>The sum of the terms in an arithmetic sequence is called an arithmetic series.<\/li>\n<li>The sum of the first [latex]n[\/latex] terms of an arithmetic series can be found using a formula.<\/li>\n<li>The sum of the terms in a geometric sequence is called a geometric series.<\/li>\n<li>The sum of the first [latex]n[\/latex] terms of a geometric series can be found using a formula.<\/li>\n<li>The sum of an infinite series exists if the series is geometric with [latex]-1<r<1[\/latex].<\/li>\n<li>If the sum of an infinite series exists, it can be found using a formula.<\/li>\n<li>An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137726792\" class=\"definition\">\n<dt>annuity<\/dt>\n<dd id=\"fs-id1165137726797\">an investment in which the purchaser makes a sequence of periodic, equal payments<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137726801\" class=\"definition\">\n<dt>arithmetic series<\/dt>\n<dd id=\"fs-id1165137726806\">the sum of the terms in an arithmetic sequence<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137726811\" class=\"definition\">\n<dt>diverge<\/dt>\n<dd id=\"fs-id1165134031382\">a series is said to diverge if the sum is not a real number<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134031386\" class=\"definition\">\n<dt>geometric series<\/dt>\n<dd id=\"fs-id1165134031391\">the sum of the terms in a geometric sequence<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134031396\" class=\"definition\">\n<dt>index of summation<\/dt>\n<dd id=\"fs-id1165134031401\">in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137737862\" class=\"definition\">\n<dt>infinite series<\/dt>\n<dd id=\"fs-id1165137737867\">the sum of the terms in an infinite sequence<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137737872\" class=\"definition\">\n<dt>lower limit of summation<\/dt>\n<dd id=\"fs-id1165137737877\">the number used in the explicit formula to find the first term in a series<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137737881\" class=\"definition\">\n<dt>nth partial sum<\/dt>\n<dd id=\"fs-id1165135471113\">the sum of the first [latex]n[\/latex] terms of a sequence<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135471124\" class=\"definition\">\n<dt>series<\/dt>\n<dd id=\"fs-id1165135471129\">the sum of the terms in a sequence<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135471133\" class=\"definition\">\n<dt>summation notation<\/dt>\n<dd id=\"fs-id1165135471138\">a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137762744\" class=\"definition\">\n<dt>upper limit of summation<\/dt>\n<dd id=\"fs-id1165137762749\">the number used in the explicit formula to find the last term in a series<\/dd>\n<\/dl>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-14807\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-14807","chapter","type-chapter","status-publish","hentry"],"part":14758,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14807","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14807\/revisions"}],"predecessor-version":[{"id":16188,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14807\/revisions\/16188"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/14758"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/14807\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=14807"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=14807"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=14807"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=14807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}