{"id":2070,"date":"2015-11-12T18:30:42","date_gmt":"2015-11-12T18:30:42","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=2070"},"modified":"2015-11-12T18:30:42","modified_gmt":"2015-11-12T18:30:42","slug":"using-the-formula-for-geometric-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/using-the-formula-for-geometric-series\/","title":{"raw":"Using the Formula for Geometric Series","rendered":"Using the Formula for Geometric Series"},"content":{"raw":"<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\n<\/p><div style=\"text-align: center;\">[latex]{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}[\/latex].<\/div>\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].\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>\nNext, we subtract this equation from the original equation.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\\\ \\frac{\\begin{array}{l}\\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\\hfill \\\\ -r{S}_{n}=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right)\\hfill \\end{array}}{\\left(1-r\\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\\end{array}[\/latex]<\/div>\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].\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>\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\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><li>Identify [latex]{a}_{1},r,\\text{and}n[\/latex].<\/li>\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>\n\t<li>Simplify to find [latex]{S}_{n}[\/latex].<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4: Finding the First <em>n<\/em> Terms of a Geometric Series<\/h3>\nUse the formula to find the indicated partial sum of each geometric series.\n<ol><li>[latex]{S}_{11}[\/latex] for the series [latex]\\text{ 8 + -4 + 2 + }\\dots [\/latex]<\/li>\n\t<li>[latex]\\underset{6}{\\overset{k=1}{{\\sum }^{\\text{ }}}}3\\cdot {2}^{k}[\/latex]<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<ol><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>\nSubstitute values for [latex]{a}_{1}, r, \\text{and} n[\/latex] into the formula and simplify.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{11}=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336\\hfill \\end{array}[\/latex]<\/div><\/li>\n\t<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>\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].\n\nSubstitute values for [latex]{a}_{1},r[\/latex], and [latex]n[\/latex] into the formula, and simplify.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378\\hfill \\end{array}[\/latex]<\/div><\/li>\n<\/ol><\/div>\nUse the formula to find the indicated partial sum of each geometric series.\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 6<\/h3>\n[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots [\/latex]\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 7<\/h3>\n[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Solving an Application Problem with a Geometric Series<\/h3>\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.\n\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{5}=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03\\hfill \\end{array}[\/latex]<\/div>\nHe will have earned a total of $138,099.03 by the end of 5 years.\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 8<\/h3>\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?\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<h2>Using the Formula for the Sum of an Infinite Geometric Series<\/h2>\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].\n\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>.\n<h2>Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h2>\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:\n<div style=\"text-align: center;\">[latex]1+0.2+0.04+0.008+0.0016+..[\/latex].<\/div>\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&lt;r&lt;1[\/latex] approach 0; the sum of a geometric series is defined when [latex]-1&lt;r&lt;1[\/latex].\n<div class=\"textbox\">\n<h3>A General Note: Determining Whether the Sum of an Infinite Geometric Series is Defined<\/h3>\nThe sum of an infinite series is defined if the series is geometric and [latex]-1&lt;r&lt;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><li>Find the ratio of the second term to the first term.<\/li>\n\t<li>Find the ratio of the third term to the second term.<\/li>\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>\n\t<li>If a common ratio, [latex]r[\/latex], was found in step 3, check to see if [latex]-1&lt;r&lt;1[\/latex] . If so, the sum is defined. If not, the sum is not defined.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 6: Determining Whether the Sum of an Infinite Series is Defined<\/h3>\nDetermine whether the sum of each infinite series is defined.\n<ol><li>[latex]\\text{12 + 8 + 4 + }\\dots [\/latex]<\/li>\n\t<li>[latex]\\frac{3}{4}+\\frac{1}{2}+\\frac{1}{3}+..[\/latex].<\/li>\n\t<li>[latex]\\sum _{k=1}^{\\infty }27\\cdot {\\left(\\frac{1}{3}\\right)}^{k}[\/latex]<\/li>\n\t<li>[latex]\\sum _{k=1}^{\\infty }5k[\/latex]<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<ol><li>The ratio of the second term to the first is [latex]\\frac{\\text{2}}{\\text{3}}[\/latex],\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\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>\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>\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>\n<\/ol><\/div>\nDetermine whether the sum of the infinite series is defined.\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 9<\/h3>\n[latex]\\frac{1}{3}+\\frac{1}{2}+\\frac{3}{4}+\\frac{9}{8}+..[\/latex].\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 10<\/h3>\n[latex]24+\\left(-12\\right)+6+\\left(-3\\right)+..[\/latex].\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 11<\/h3>\n[latex]\\sum _{k=1}^{\\infty }15\\cdot {\\left(-0.3\\right)}^{k}[\/latex]\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a>\n\n<\/div>","rendered":"<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\n<\/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{array}{l}\\\\ \\frac{\\begin{array}{l}\\text{ }{S}_{n}={a}_{1}+r{a}_{1}+{r}^{2}{a}_{1}+...+{r}^{n - 1}{a}_{1}\\hfill \\\\ -r{S}_{n}=-\\left(r{a}_{1}+{r}^{2}{a}_{1}+{r}^{3}{a}_{1}+...+{r}^{n}{a}_{1}\\right)\\hfill \\end{array}}{\\left(1-r\\right){S}_{n}={a}_{1}-{r}^{n}{a}_{1}}\\end{array}[\/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]\\underset{6}{\\overset{k=1}{{\\sum }^{\\text{ }}}}3\\cdot {2}^{k}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{11}=\\frac{8\\left(1-{\\left(-\\frac{1}{2}\\right)}^{11}\\right)}{1-\\left(-\\frac{1}{2}\\right)}\\approx 5.336\\hfill \\end{array}[\/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{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{6}=\\frac{6\\left(1-{2}^{6}\\right)}{1 - 2}=378\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<\/ol>\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 6<\/h3>\n<p>[latex]{S}_{20}[\/latex] for the series [latex]\\text{ 1,000 + 500 + 250 + }\\dots[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 7<\/h3>\n<p>[latex]\\sum _{k=1}^{8}{3}^{k}[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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<div style=\"text-align: center;\">[latex]\\begin{array}{l}{S}_{n}=\\frac{{a}_{1}\\left(1-{r}^{n}\\right)}{1-r}\\hfill \\\\ {S}_{5}=\\frac{26\\text{,}750\\left(1-{1.016}^{5}\\right)}{1 - 1.016}\\approx 138\\text{,}099.03\\hfill \\end{array}[\/latex]<\/div>\n<p>He will have earned a total of $138,099.03 by the end of 5 years.<\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 8<\/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<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/p>\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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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<p>Determine whether the sum of the infinite series is defined.<\/p>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 9<\/h3>\n<p>[latex]\\frac{1}{3}+\\frac{1}{2}+\\frac{3}{4}+\\frac{9}{8}+..[\/latex].<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 10<\/h3>\n<p>[latex]24+\\left(-12\\right)+6+\\left(-3\\right)+..[\/latex].<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 11<\/h3>\n<p>[latex]\\sum _{k=1}^{\\infty }15\\cdot {\\left(-0.3\\right)}^{k}[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-31\/\" target=\"_blank\">Solution<\/a><\/p>\n<\/div>\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-2070\">\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":276,"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-2070","chapter","type-chapter","status-publish","hentry"],"part":2065,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2070","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2070\/revisions"}],"predecessor-version":[{"id":2158,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2070\/revisions\/2158"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2065"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2070\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=2070"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2070"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2070"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=2070"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}