{"id":1724,"date":"2021-07-23T20:33:29","date_gmt":"2021-07-23T20:33:29","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1724"},"modified":"2022-03-21T23:07:02","modified_gmt":"2022-03-21T23:07:02","slug":"choosing-a-convergence-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/choosing-a-convergence-test\/","title":{"raw":"Choosing a Convergence Test","rendered":"Choosing a Convergence Test"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe a strategy for testing the convergence of a given series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169736592468\">At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.<\/p>\r\n\r\n<div id=\"fs-id1169736592473\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 data-type=\"title\">Problem-Solving Strategy: Choosing a Convergence Test for a Series<\/h3>\r\n<p id=\"fs-id1169736592480\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. In the steps below, we outline a strategy for determining whether the series converges.<\/p>\r\n\r\n<ol id=\"fs-id1169736592510\" type=\"1\">\r\n \t<li>Is [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a [latex]p-\\text{series}[\/latex] or geometric series? If so, check the power [latex]p[\/latex] or the ratio [latex]r[\/latex] to determine if the series converges.<\/li>\r\n \t<li>Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. If we are interested in absolute convergence, proceed to step [latex]3[\/latex], considering the series of absolute values [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex].<\/li>\r\n \t<li>Is the series similar to a [latex]p-\\text{series}[\/latex] or geometric series? If so, try the comparison test or limit comparison test.<\/li>\r\n \t<li>Do the terms in the series contain a factorial or power? If the terms are powers such that [latex]{a}_{n}={b}_{n}^{n}[\/latex], try the root test first. Otherwise, try the ratio test first.<\/li>\r\n \t<li>Use the divergence test. If this test does not provide any information, try the integral test.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169736685638\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox tryit\">\r\n<h3>Media<\/h3>\r\n<p id=\"fs-id1169736685641\">Visit <a href=\"https:\/\/www.csusm.edu\/mathlab\/documents\/seriesconvergediverge.pdf\" target=\"_blank\" rel=\"noopener\">this website for more information on testing series for convergence<\/a>, plus general information on sequences and series.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169736685651\" data-type=\"example\">\r\n<div id=\"fs-id1169736685653\" data-type=\"exercise\">\r\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Using Convergence Tests<\/h3>\r\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\r\n<p id=\"fs-id1169736685660\">For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169736685666\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}\\left(3n+1\\right)}{n\\text{!}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{{n}^{3}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{3}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169736852464\" data-type=\"solution\">\r\n<ol id=\"fs-id1169736852466\" type=\"a\">\r\n \t<li>Step 1. The series is not a [latex]p-\\text{series}[\/latex] or geometric series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. The series is not alternating.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. For large values of [latex]n[\/latex], we approximate the series by the expression<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736852496\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\\approx \\frac{{n}^{2}}{{n}^{3}}=\\frac{1}{n}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, it seems reasonable to apply the comparison test or limit comparison test using the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Using the limit comparison test, we see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736843074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({n}^{2}+2n\\right)}{\\left({n}^{3}+3{n}^{2}+1\\right)}}{\\frac{1}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, this series diverges as well.<\/li>\r\n \t<li>Step 1.The series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. The series is alternating. Since we are interested in absolute convergence, consider the series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736747084\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3n}{\\left(n+1\\right)\\text{!}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. The series is not similar to a <em data-effect=\"italics\">p<\/em>-series or geometric series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. Since each term contains a factorial, apply the ratio test. We see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736643942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left(3\\left(n+1\\right)\\right)}{\\left(n+1\\right)\\text{!}}}{\\frac{\\left(3n+1\\right)}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{3n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\left(3n+1\\right)}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.<\/li>\r\n \t<li>Step 1. The series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. It is not an alternating series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.<span data-type=\"newline\">\r\n<\/span>\r\nStep 5. To apply the divergence test, we calculate that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736843793\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{e}^{n}}{{n}^{3}}=\\infty [\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, by the divergence test, the series diverges.<\/li>\r\n \t<li>Step 1. This series is not a familiar series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 2. It is not an alternating series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\">\r\n<\/span>\r\nStep 4. Since each term is a power of [latex]n[\/latex], we can apply the root test. Since<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739077365\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{{\\left(\\frac{3}{n+1}\\right)}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3}{n+1}=0[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nby the root test, we conclude that the series converges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1169739077449\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169739077452\" data-type=\"exercise\">\r\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\r\n<p id=\"fs-id1169739077456\">For the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{3}^{n}+n}[\/latex], determine which convergence test is the best to use and explain why.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1169739041818\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739041825\">The series is similar to the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1169739041763\" data-type=\"solution\">\r\n<p id=\"fs-id1169739041765\">The comparison test because [latex]\\frac{{2}^{n}}{\\left({3}^{n}+n\\right)}&lt;\\frac{{2}^{n}}{{3}^{n}}[\/latex] for all positive integers [latex]n[\/latex]. The limit comparison test could also be used.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YpHtcijxlzw?controls=0&amp;start=813&amp;end=789&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/MixedConvergenceTests813to789_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"Mixed Convergence Tests\" here (opens in new window)<\/a>.\r\n<p id=\"fs-id1169739027066\">In the following table, we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] to have nonnegative terms, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] has negative terms, these tests can be applied to [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] to test for absolute convergence.<\/p>\r\n\r\n<table id=\"fs-id1169739027151\" style=\"height: 788px;\" summary=\"This is a table with three columns and many rows describing the different series or test, conclusions, and comments in each column, respectively. The first test\/series is the divergence test: for any series written as the sum from n = 1 to infinity of a_n, evaluate the limit as n goes to infinity of a + n. If the limit is 0, the test is inconclusive. If the limit is not equal to 0, the series diverges. Note \u2013 this test cannot prove convergence of a series. The second test\/series is the geometric series: the sum from n = 1 to infinity of a r^(n - 1). If the absolute value of r is less than 1, the series converges to a\/(1 - r). If the absolute value of r is greater than or equal to 1, the series diverges. Note \u2013 any geometric series can be reindexed to be written in the form a + a r + a r^2 + \u2026, where a is the initial term and r is the ratio. The third test\/series is the p-series: the sum from n = 1 to infinity of 1\/n^p. If p is greater than 1, the series converges. If p is less than or equal to 1, the series diverges. Note \u2014 for p = 1, we have the harmonic series: the sum from n = 1 to infinity of 1\/n. The fourth test\/series is the comparison test: for the sum from n = 1 to infinity of a_n with nonnegative terms, compare with a known series \u2013 the sum from n = 1 to infinity of b_n. If a_n is less than or equal to b_n for all n greater than or equal to N and the sum from n = 1 to infinity of b_n converges, then the sum from n = 1 to infinity of a_n converges. If a_n is greater than or equal to b_n for all n greater than or equal to N and the sum from n = 1 to infinity of b_n diverges, then the sum from n = 1 to infinity a_n diverges. Note \u2014 this is typically used for a series similar to a geometric or p-series. It can sometimes be difficult to find an appropriate series. The fifth test\/series is the limit comparison test: for the sum from n = 1 to infinity of a_n with positive terms, compare with a series \u2013 the sum from n = 1 to infinity of b_n by evaluating L = the limit as n goes to infinity of a_n \/ b_n. If L is a real number and L does not equal zero, then the sum from n=1 to infinity of a_n and the sum from n = 1 to infinity of b_n both converge or both diverge. If L = 0 and the sum from n = 1 to infinity of b_n converges, then the sum from n = 1 to infinity of a_n converges. If L = infinity and the sum from n = 1 to infinity of b_n diverges, then the sum from n = 1 to infinity of a_n diverges. Note \u2014 this is typically used for a series similar to a geometric of p-series. It is often easier to apply than the comparison test. The sixth test\/series is the integral test: if there exists a positive, continuous, decreasing function f such that a_n = f(n) for all n greater than or equal to N, evaluate the integral from N to infinity of f(x) d x. Both the integral from N to infinity of f(x)f x and the sum from n = 1 to infinity will converge or diverge. Note \u2014 this is limited to those series for which the corresponding function f can be easily integrated. The seventh test\/series is the alternating series: the sum from n = 1 to infinity of (-1)^(n + 1) * b_n or the sum from n = 1 to infinity of (-1)^n * b_n. If b_(n + 1) is less than or equal to b_n for all n greater than or equal to 1 and b_n goes to zero, then the series converges. Note \u2014 this only applies to alternating series. The eighth test\/series is the ratio test: for any series \u2013 the sum from n = 1 to infinity of a_n \u2013 with nonzero terms, let p equal the limit as n goes to infinity of the absolute value of a_(n + 1)\/a_n. If 0 is less than or equal to p which is less than 1, the series converges absolutely. If p is greater than 1 or p equals infinity, then the series diverges. If p equals 1, the test is inconclusive. Note \u2014 this is often used for series involving factorials or exponentials. The ninth test\/series is the root test: for any series \u2013 the sum from n = 1 to infinity of a_n \u2013 let p equal the limit as n goes to infinity of the n-th root of the absolute value of a_n. If 0 is less than or equal to p which is less than 1, the series converges absolutely. If p is less than 1 or p equals infinity the series diverges. If p equals 1, the test is inconclusive. Note \u2014 often used for series where the absolute value of a_n = b_n to the power of n.\"><caption><span data-type=\"title\">Summary of Convergence Tests<\/span><\/caption>\r\n<thead>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<th style=\"height: 14px; width: 327.328px;\" data-align=\"left\">Series or Test<\/th>\r\n<th style=\"height: 14px; width: 281.969px;\" data-align=\"left\">Conclusions<\/th>\r\n<th style=\"height: 14px; width: 268.203px;\" data-align=\"left\">Comments<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 28px;\">\r\n<td style=\"height: 56px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Divergence Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex].<\/td>\r\n<td style=\"height: 28px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex], the test is inconclusive.<\/td>\r\n<td style=\"height: 56px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">This test cannot prove convergence of a series.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 28px;\">\r\n<td style=\"height: 28px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}\\ne 0[\/latex], the series diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 42px;\">\r\n<td style=\"height: 56px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Geometric Series<\/strong><span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]<\/td>\r\n<td style=\"height: 42px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]|r|&lt;1[\/latex], the series converges to<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\frac{a}{\\left(1-r\\right)}[\/latex].<\/td>\r\n<td style=\"height: 56px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">Any geometric series can be reindexed to be written in the form [latex]a+ar+a{r}^{2}+\\cdots [\/latex], where [latex]a[\/latex] is the initial term and [latex]r[\/latex] is the ratio.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]|r|\\ge 1[\/latex], the series diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 21px;\">\r\n<td style=\"height: 42px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\"><em data-effect=\"italics\">p<\/em>-Series<\/strong><span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/td>\r\n<td style=\"height: 21px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]p&gt;1[\/latex], the series converges.<\/td>\r\n<td style=\"height: 42px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">For [latex]p=1[\/latex], we have the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex].<\/td>\r\n<\/tr>\r\n<tr style=\"height: 21px;\">\r\n<td style=\"height: 21px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]p\\le 1[\/latex], the series diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 70px;\">\r\n<td style=\"height: 140px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Comparison Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nFor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with nonnegative terms, compare with a known series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/td>\r\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/td>\r\n<td style=\"height: 140px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">Typically used for a series similar to a geometric or [latex]p[\/latex] -series. It can sometimes be difficult to find an appropriate series.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 70px;\">\r\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{a}_{n}\\ge {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 70px;\">\r\n<td style=\"height: 182px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Limit Comparison Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nFor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with positive terms, compare with a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] by evaluating<span data-type=\"newline\">\r\n<\/span>\r\n[latex]L=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/td>\r\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L[\/latex] is a real number and [latex]L\\ne 0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] both converge or both diverge.<\/td>\r\n<td style=\"height: 182px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Typically used for a series similar to a geometric or [latex]p[\/latex] -series. Often easier to apply than the comparison test.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 56px;\">\r\n<td style=\"height: 56px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 56px;\">\r\n<td style=\"height: 56px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L=\\infty [\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 84px;\">\r\n<td style=\"height: 84px; width: 327.328px;\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Integral Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nIf there exists a positive, continuous, decreasing function [latex]f[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all [latex]n\\ge N[\/latex], evaluate [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/td>\r\n<td style=\"height: 84px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">[latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] both converge or both diverge.<\/td>\r\n<td style=\"height: 84px; width: 268.203px;\" data-valign=\"top\" data-align=\"left\">Limited to those series for which the corresponding function [latex]f[\/latex] can be easily integrated.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 70px;\">\r\n<td style=\"height: 70px; width: 327.328px;\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Alternating Series<\/strong><span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/td>\r\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge 1[\/latex] and [latex]{b}_{n}\\to 0[\/latex], then the series converges.<\/td>\r\n<td style=\"height: 70px; width: 268.203px;\" data-valign=\"top\" data-align=\"left\">Only applies to alternating series.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 29px;\">\r\n<td style=\"height: 72px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Ratio Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with nonzero terms, let<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|[\/latex].<\/td>\r\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]0\\le \\rho &lt;1[\/latex], the series converges absolutely.<\/td>\r\n<td style=\"height: 72px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Often used for series involving factorials or exponentials.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 29px;\">\r\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho &gt;1\\text{or}\\rho =\\infty [\/latex], the series diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho =1[\/latex], the test is inconclusive.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 29px;\">\r\n<td style=\"height: 72px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Root Test<\/strong><span data-type=\"newline\">\r\n<\/span>\r\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], let<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}[\/latex].<\/td>\r\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]0\\le \\rho &lt;1[\/latex], the series converges absolutely.<\/td>\r\n<td style=\"height: 72px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Often used for series where [latex]|{a}_{n}|={b}_{n}^{n}[\/latex].<\/td>\r\n<\/tr>\r\n<tr style=\"height: 29px;\">\r\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho &gt;1\\text{or}\\rho =\\infty [\/latex], the series diverges.<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho =1[\/latex], the test is inconclusive.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1169736582702\" class=\"project\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3 data-type=\"title\">Activity: Series Converging to [latex]\\pi [\/latex] and [latex]\\frac{1}{\\pi} [\/latex]<\/h3>\r\n<p id=\"fs-id1169736582723\">Dozens of series exist that converge to [latex]\\pi [\/latex] or an algebraic expression containing [latex]\\pi [\/latex]. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of [latex]\\pi [\/latex] in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.<\/p>\r\n\r\n<ol id=\"fs-id1169736707593\" type=\"1\">\r\n \t<li>The series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736707603\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\pi =4\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{2n - 1}=4-\\frac{4}{3}+\\frac{4}{5}-\\frac{4}{7}+\\frac{4}{9}-\\cdots [\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwas discovered by Gregory and Leibniz in the late [latex]1600\\text{s}\\text{.}[\/latex] This result follows from the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]. We will discuss this series in the next chapter.<span data-type=\"newline\">\r\n<\/span>\r\n<ol id=\"fs-id1169739171821\" type=\"a\">\r\n \t<li>Prove that this series converges.<\/li>\r\n \t<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=10,20,50,100[\/latex].<\/li>\r\n \t<li>Use the remainder estimate for alternating series to get a bound on the error [latex]{R}_{n}[\/latex].<\/li>\r\n \t<li>What is the smallest value of [latex]N[\/latex] that guarantees [latex]|{R}_{N}|&lt;0.01\\text{?}[\/latex] Evaluate [latex]{S}_{N}[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739171920\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\pi &amp; =6 {\\displaystyle\\sum _{n=0}^{\\infty }} \\frac{\\left(2n\\right)\\text{!}}{{2}^{4n+1}{\\left(n\\text{!}\\right)}^{2}\\left(2n+1\\right)}\\hfill \\\\ &amp; =6\\left(\\frac{1}{2}+\\frac{1}{2\\cdot 3}{\\left(\\frac{1}{2}\\right)}^{3}+\\frac{1\\cdot 3}{2\\cdot 4\\cdot 5}\\cdot {\\left(\\frac{1}{2}\\right)}^{5}+\\frac{1\\cdot 3\\cdot 5}{2\\cdot 4\\cdot 6\\cdot 7}{\\left(\\frac{1}{2}\\right)}^{7}+\\cdots \\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nhas been attributed to Newton in the late [latex]1600\\text{s}\\text{.}[\/latex] The proof of this result uses the Maclaurin series for [latex]f\\left(x\\right)={\\sin}^{-1}x[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<ol id=\"fs-id1169736851958\" type=\"a\">\r\n \t<li>Prove that the series converges.<\/li>\r\n \t<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=5,10,20[\/latex].<\/li>\r\n \t<li>Compare [latex]{S}_{n}[\/latex] to [latex]\\pi [\/latex] for [latex]n=5,10,20[\/latex] and discuss the number of correct decimal places.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>The series<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738920677\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\pi }=\\frac{\\sqrt{8}}{9801}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\left(4n\\right)\\text{!}\\left(1103+26390n\\right)}{{\\left(n\\text{!}\\right)}^{4}{396}^{4n}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwas discovered by <span class=\"no-emphasis\" data-type=\"term\">Ramanujan<\/span> in the early [latex]1900\\text{s}\\text{.}[\/latex] William Gosper, Jr., used this series to calculate [latex]\\pi [\/latex] to an accuracy of more than [latex]17[\/latex] million digits in the [latex]\\text{mid-}1980\\text{s}\\text{.}[\/latex] At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for [latex]\\pi [\/latex] and [latex]\\frac{1}{\\pi} [\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<ol id=\"fs-id1169739252686\" type=\"a\">\r\n \t<li>Prove that this series converges.<\/li>\r\n \t<li>Evaluate the first term in this series. Compare this number with the value of [latex]\\pi [\/latex] from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?<\/li>\r\n \t<li>Investigate the life of Srinivasa Ramanujan [latex]\\left(1887\\text{-}1920\\right)[\/latex] and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169739252731\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe a strategy for testing the convergence of a given series<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169736592468\">At this point, we have a long list of convergence tests. However, not all tests can be used for all series. When given a series, we must determine which test is the best to use. Here is a strategy for finding the best test to apply.<\/p>\n<div id=\"fs-id1169736592473\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 data-type=\"title\">Problem-Solving Strategy: Choosing a Convergence Test for a Series<\/h3>\n<p id=\"fs-id1169736592480\">Consider a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. In the steps below, we outline a strategy for determining whether the series converges.<\/p>\n<ol id=\"fs-id1169736592510\" type=\"1\">\n<li>Is [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] a familiar series? For example, is it the harmonic series (which diverges) or the alternating harmonic series (which converges)? Is it a [latex]p-\\text{series}[\/latex] or geometric series? If so, check the power [latex]p[\/latex] or the ratio [latex]r[\/latex] to determine if the series converges.<\/li>\n<li>Is it an alternating series? Are we interested in absolute convergence or just convergence? If we are just interested in whether the series converges, apply the alternating series test. If we are interested in absolute convergence, proceed to step [latex]3[\/latex], considering the series of absolute values [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex].<\/li>\n<li>Is the series similar to a [latex]p-\\text{series}[\/latex] or geometric series? If so, try the comparison test or limit comparison test.<\/li>\n<li>Do the terms in the series contain a factorial or power? If the terms are powers such that [latex]{a}_{n}={b}_{n}^{n}[\/latex], try the root test first. Otherwise, try the ratio test first.<\/li>\n<li>Use the divergence test. If this test does not provide any information, try the integral test.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736685638\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox tryit\">\n<h3>Media<\/h3>\n<p id=\"fs-id1169736685641\">Visit <a href=\"https:\/\/www.csusm.edu\/mathlab\/documents\/seriesconvergediverge.pdf\" target=\"_blank\" rel=\"noopener\">this website for more information on testing series for convergence<\/a>, plus general information on sequences and series.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736685651\" data-type=\"example\">\n<div id=\"fs-id1169736685653\" data-type=\"exercise\">\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Using Convergence Tests<\/h3>\n<div id=\"fs-id1169736685655\" data-type=\"problem\">\n<p id=\"fs-id1169736685660\">For each of the following series, determine which convergence test is the best to use and explain why. Then determine if the series converges or diverges. If the series is an alternating series, determine whether it converges absolutely, converges conditionally, or diverges.<\/p>\n<ol id=\"fs-id1169736685666\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}\\left(3n+1\\right)}{n\\text{!}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{e}^{n}}{{n}^{3}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{3}^{n}}{{\\left(n+1\\right)}^{n}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736852464\" data-type=\"solution\">\n<ol id=\"fs-id1169736852466\" type=\"a\">\n<li>Step 1. The series is not a [latex]p-\\text{series}[\/latex] or geometric series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. The series is not alternating.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. For large values of [latex]n[\/latex], we approximate the series by the expression<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736852496\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{{n}^{2}+2n}{{n}^{3}+3{n}^{2}+1}\\approx \\frac{{n}^{2}}{{n}^{3}}=\\frac{1}{n}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, it seems reasonable to apply the comparison test or limit comparison test using the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. Using the limit comparison test, we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736843074\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({n}^{2}+2n\\right)}{\\left({n}^{3}+3{n}^{2}+1\\right)}}{\\frac{1}{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{3}+2{n}^{2}}{{n}^{3}+3{n}^{2}+1}=1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, this series diverges as well.<\/li>\n<li>Step 1.The series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. The series is alternating. Since we are interested in absolute convergence, consider the series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736747084\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{3n}{\\left(n+1\\right)\\text{!}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. The series is not similar to a <em data-effect=\"italics\">p<\/em>-series or geometric series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. Since each term contains a factorial, apply the ratio test. We see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736643942\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left(3\\left(n+1\\right)\\right)}{\\left(n+1\\right)\\text{!}}}{\\frac{\\left(3n+1\\right)}{n\\text{!}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\text{!}}\\cdot \\frac{n\\text{!}}{3n+1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3n+3}{\\left(n+1\\right)\\left(3n+1\\right)}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, this series converges, and we conclude that the original series converges absolutely, and thus converges.<\/li>\n<li>Step 1. The series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. It is not an alternating series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. There is no factorial. There is a power, but it is not an ideal situation for the root test.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 5. To apply the divergence test, we calculate that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736843793\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{e}^{n}}{{n}^{3}}=\\infty[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, by the divergence test, the series diverges.<\/li>\n<li>Step 1. This series is not a familiar series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 2. It is not an alternating series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 3. There is no obvious series with which to compare this series.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nStep 4. Since each term is a power of [latex]n[\/latex], we can apply the root test. Since<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739077365\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{{\\left(\\frac{3}{n+1}\\right)}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{3}{n+1}=0[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nby the root test, we conclude that the series converges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169739077449\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169739077452\" data-type=\"exercise\">\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169739077454\" data-type=\"problem\">\n<p id=\"fs-id1169739077456\">For the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}}{{3}^{n}+n}[\/latex], determine which convergence test is the best to use and explain why.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Hint<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739041818\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739041825\">The series is similar to the geometric series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739041763\" data-type=\"solution\">\n<p id=\"fs-id1169739041765\">The comparison test because [latex]\\frac{{2}^{n}}{\\left({3}^{n}+n\\right)}<\\frac{{2}^{n}}{{3}^{n}}[\/latex] for all positive integers [latex]n[\/latex]. The limit comparison test could also be used.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/YpHtcijxlzw?controls=0&amp;start=813&amp;end=789&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/MixedConvergenceTests813to789_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;Mixed Convergence Tests&#8221; here (opens in new window)<\/a>.<\/p>\n<p id=\"fs-id1169739027066\">In the following table, we summarize the convergence tests and when each can be applied. Note that while the comparison test, limit comparison test, and integral test require the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] to have nonnegative terms, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] has negative terms, these tests can be applied to [latex]\\displaystyle\\sum _{n=1}^{\\infty }|{a}_{n}|[\/latex] to test for absolute convergence.<\/p>\n<table id=\"fs-id1169739027151\" style=\"height: 788px;\" summary=\"This is a table with three columns and many rows describing the different series or test, conclusions, and comments in each column, respectively. The first test\/series is the divergence test: for any series written as the sum from n = 1 to infinity of a_n, evaluate the limit as n goes to infinity of a + n. If the limit is 0, the test is inconclusive. If the limit is not equal to 0, the series diverges. Note \u2013 this test cannot prove convergence of a series. The second test\/series is the geometric series: the sum from n = 1 to infinity of a r^(n - 1). If the absolute value of r is less than 1, the series converges to a\/(1 - r). If the absolute value of r is greater than or equal to 1, the series diverges. Note \u2013 any geometric series can be reindexed to be written in the form a + a r + a r^2 + \u2026, where a is the initial term and r is the ratio. The third test\/series is the p-series: the sum from n = 1 to infinity of 1\/n^p. If p is greater than 1, the series converges. If p is less than or equal to 1, the series diverges. Note \u2014 for p = 1, we have the harmonic series: the sum from n = 1 to infinity of 1\/n. The fourth test\/series is the comparison test: for the sum from n = 1 to infinity of a_n with nonnegative terms, compare with a known series \u2013 the sum from n = 1 to infinity of b_n. If a_n is less than or equal to b_n for all n greater than or equal to N and the sum from n = 1 to infinity of b_n converges, then the sum from n = 1 to infinity of a_n converges. If a_n is greater than or equal to b_n for all n greater than or equal to N and the sum from n = 1 to infinity of b_n diverges, then the sum from n = 1 to infinity a_n diverges. Note \u2014 this is typically used for a series similar to a geometric or p-series. It can sometimes be difficult to find an appropriate series. The fifth test\/series is the limit comparison test: for the sum from n = 1 to infinity of a_n with positive terms, compare with a series \u2013 the sum from n = 1 to infinity of b_n by evaluating L = the limit as n goes to infinity of a_n \/ b_n. If L is a real number and L does not equal zero, then the sum from n=1 to infinity of a_n and the sum from n = 1 to infinity of b_n both converge or both diverge. If L = 0 and the sum from n = 1 to infinity of b_n converges, then the sum from n = 1 to infinity of a_n converges. If L = infinity and the sum from n = 1 to infinity of b_n diverges, then the sum from n = 1 to infinity of a_n diverges. Note \u2014 this is typically used for a series similar to a geometric of p-series. It is often easier to apply than the comparison test. The sixth test\/series is the integral test: if there exists a positive, continuous, decreasing function f such that a_n = f(n) for all n greater than or equal to N, evaluate the integral from N to infinity of f(x) d x. Both the integral from N to infinity of f(x)f x and the sum from n = 1 to infinity will converge or diverge. Note \u2014 this is limited to those series for which the corresponding function f can be easily integrated. The seventh test\/series is the alternating series: the sum from n = 1 to infinity of (-1)^(n + 1) * b_n or the sum from n = 1 to infinity of (-1)^n * b_n. If b_(n + 1) is less than or equal to b_n for all n greater than or equal to 1 and b_n goes to zero, then the series converges. Note \u2014 this only applies to alternating series. The eighth test\/series is the ratio test: for any series \u2013 the sum from n = 1 to infinity of a_n \u2013 with nonzero terms, let p equal the limit as n goes to infinity of the absolute value of a_(n + 1)\/a_n. If 0 is less than or equal to p which is less than 1, the series converges absolutely. If p is greater than 1 or p equals infinity, then the series diverges. If p equals 1, the test is inconclusive. Note \u2014 this is often used for series involving factorials or exponentials. The ninth test\/series is the root test: for any series \u2013 the sum from n = 1 to infinity of a_n \u2013 let p equal the limit as n goes to infinity of the n-th root of the absolute value of a_n. If 0 is less than or equal to p which is less than 1, the series converges absolutely. If p is less than 1 or p equals infinity the series diverges. If p equals 1, the test is inconclusive. Note \u2014 often used for series where the absolute value of a_n = b_n to the power of n.\">\n<caption><span data-type=\"title\">Summary of Convergence Tests<\/span><\/caption>\n<thead>\n<tr style=\"height: 14px;\" valign=\"top\">\n<th style=\"height: 14px; width: 327.328px;\" data-align=\"left\">Series or Test<\/th>\n<th style=\"height: 14px; width: 281.969px;\" data-align=\"left\">Conclusions<\/th>\n<th style=\"height: 14px; width: 268.203px;\" data-align=\"left\">Comments<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 28px;\">\n<td style=\"height: 56px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Divergence Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}[\/latex].<\/td>\n<td style=\"height: 28px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}=0[\/latex], the test is inconclusive.<\/td>\n<td style=\"height: 56px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">This test cannot prove convergence of a series.<\/td>\n<\/tr>\n<tr style=\"height: 28px;\">\n<td style=\"height: 28px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\underset{n\\to \\infty }{\\text{lim}}{a}_{n}\\ne 0[\/latex], the series diverges.<\/td>\n<\/tr>\n<tr style=\"height: 42px;\">\n<td style=\"height: 56px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Geometric Series<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }a{r}^{n - 1}[\/latex]<\/td>\n<td style=\"height: 42px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]|r|<1[\/latex], the series converges to<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\frac{a}{\\left(1-r\\right)}[\/latex].<\/td>\n<td style=\"height: 56px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">Any geometric series can be reindexed to be written in the form [latex]a+ar+a{r}^{2}+\\cdots[\/latex], where [latex]a[\/latex] is the initial term and [latex]r[\/latex] is the ratio.<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]|r|\\ge 1[\/latex], the series diverges.<\/td>\n<\/tr>\n<tr style=\"height: 21px;\">\n<td style=\"height: 42px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\"><em data-effect=\"italics\">p<\/em>-Series<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/td>\n<td style=\"height: 21px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]p>1[\/latex], the series converges.<\/td>\n<td style=\"height: 42px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">For [latex]p=1[\/latex], we have the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex].<\/td>\n<\/tr>\n<tr style=\"height: 21px;\">\n<td style=\"height: 21px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]p\\le 1[\/latex], the series diverges.<\/td>\n<\/tr>\n<tr style=\"height: 70px;\">\n<td style=\"height: 140px; width: 327.328px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Comparison Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with nonnegative terms, compare with a known series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex].<\/td>\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{a}_{n}\\le {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/td>\n<td style=\"height: 140px; width: 268.203px;\" rowspan=\"2\" data-valign=\"top\" data-align=\"left\">Typically used for a series similar to a geometric or [latex]p[\/latex] -series. It can sometimes be difficult to find an appropriate series.<\/td>\n<\/tr>\n<tr style=\"height: 70px;\">\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{a}_{n}\\ge {b}_{n}[\/latex] for all [latex]n\\ge N[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/td>\n<\/tr>\n<tr style=\"height: 70px;\">\n<td style=\"height: 182px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Limit Comparison Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with positive terms, compare with a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] by evaluating<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]L=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/td>\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L[\/latex] is a real number and [latex]L\\ne 0[\/latex], then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] both converge or both diverge.<\/td>\n<td style=\"height: 182px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Typically used for a series similar to a geometric or [latex]p[\/latex] -series. Often easier to apply than the comparison test.<\/td>\n<\/tr>\n<tr style=\"height: 56px;\">\n<td style=\"height: 56px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/td>\n<\/tr>\n<tr style=\"height: 56px;\">\n<td style=\"height: 56px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]L=\\infty[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/td>\n<\/tr>\n<tr style=\"height: 84px;\">\n<td style=\"height: 84px; width: 327.328px;\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Integral Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIf there exists a positive, continuous, decreasing function [latex]f[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all [latex]n\\ge N[\/latex], evaluate [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/td>\n<td style=\"height: 84px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">[latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] both converge or both diverge.<\/td>\n<td style=\"height: 84px; width: 268.203px;\" data-valign=\"top\" data-align=\"left\">Limited to those series for which the corresponding function [latex]f[\/latex] can be easily integrated.<\/td>\n<\/tr>\n<tr style=\"height: 70px;\">\n<td style=\"height: 70px; width: 327.328px;\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Alternating Series<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n+1}{b}_{n}\\text{or}\\displaystyle\\sum _{n=1}^{\\infty }{\\left(-1\\right)}^{n}{b}_{n}[\/latex]<\/td>\n<td style=\"height: 70px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]{b}_{n+1}\\le {b}_{n}[\/latex] for all [latex]n\\ge 1[\/latex] and [latex]{b}_{n}\\to 0[\/latex], then the series converges.<\/td>\n<td style=\"height: 70px; width: 268.203px;\" data-valign=\"top\" data-align=\"left\">Only applies to alternating series.<\/td>\n<\/tr>\n<tr style=\"height: 29px;\">\n<td style=\"height: 72px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Ratio Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] with nonzero terms, let<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}|\\frac{{a}_{n+1}}{{a}_{n}}|[\/latex].<\/td>\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]0\\le \\rho <1[\/latex], the series converges absolutely.<\/td>\n<td style=\"height: 72px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Often used for series involving factorials or exponentials.<\/td>\n<\/tr>\n<tr style=\"height: 29px;\">\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho >1\\text{or}\\rho =\\infty[\/latex], the series diverges.<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho =1[\/latex], the test is inconclusive.<\/td>\n<\/tr>\n<tr style=\"height: 29px;\">\n<td style=\"height: 72px; width: 327.328px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Root Test<\/strong><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFor any series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], let<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\rho =\\underset{n\\to \\infty }{\\text{lim}}\\sqrt[n]{|{a}_{n}|}[\/latex].<\/td>\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]0\\le \\rho <1[\/latex], the series converges absolutely.<\/td>\n<td style=\"height: 72px; width: 268.203px;\" rowspan=\"3\" data-valign=\"top\" data-align=\"left\">Often used for series where [latex]|{a}_{n}|={b}_{n}^{n}[\/latex].<\/td>\n<\/tr>\n<tr style=\"height: 29px;\">\n<td style=\"height: 29px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho >1\\text{or}\\rho =\\infty[\/latex], the series diverges.<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"height: 14px; width: 281.969px;\" data-valign=\"top\" data-align=\"left\">If [latex]\\rho =1[\/latex], the test is inconclusive.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1169736582702\" class=\"project\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3 data-type=\"title\">Activity: Series Converging to [latex]\\pi[\/latex] and [latex]\\frac{1}{\\pi}[\/latex]<\/h3>\n<p id=\"fs-id1169736582723\">Dozens of series exist that converge to [latex]\\pi[\/latex] or an algebraic expression containing [latex]\\pi[\/latex]. Here we look at several examples and compare their rates of convergence. By rate of convergence, we mean the number of terms necessary for a partial sum to be within a certain amount of the actual value. The series representations of [latex]\\pi[\/latex] in the first two examples can be explained using Maclaurin series, which are discussed in the next chapter. The third example relies on material beyond the scope of this text.<\/p>\n<ol id=\"fs-id1169736707593\" type=\"1\">\n<li>The series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736707603\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\pi =4\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{\\left(-1\\right)}^{n+1}}{2n - 1}=4-\\frac{4}{3}+\\frac{4}{5}-\\frac{4}{7}+\\frac{4}{9}-\\cdots[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwas discovered by Gregory and Leibniz in the late [latex]1600\\text{s}\\text{.}[\/latex] This result follows from the Maclaurin series for [latex]f\\left(x\\right)={\\tan}^{-1}x[\/latex]. We will discuss this series in the next chapter.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1169739171821\" type=\"a\">\n<li>Prove that this series converges.<\/li>\n<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=10,20,50,100[\/latex].<\/li>\n<li>Use the remainder estimate for alternating series to get a bound on the error [latex]{R}_{n}[\/latex].<\/li>\n<li>What is the smallest value of [latex]N[\/latex] that guarantees [latex]|{R}_{N}|<0.01\\text{?}[\/latex] Evaluate [latex]{S}_{N}[\/latex].<\/li>\n<\/ol>\n<\/li>\n<li>The series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739171920\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\pi & =6 {\\displaystyle\\sum _{n=0}^{\\infty }} \\frac{\\left(2n\\right)\\text{!}}{{2}^{4n+1}{\\left(n\\text{!}\\right)}^{2}\\left(2n+1\\right)}\\hfill \\\\ & =6\\left(\\frac{1}{2}+\\frac{1}{2\\cdot 3}{\\left(\\frac{1}{2}\\right)}^{3}+\\frac{1\\cdot 3}{2\\cdot 4\\cdot 5}\\cdot {\\left(\\frac{1}{2}\\right)}^{5}+\\frac{1\\cdot 3\\cdot 5}{2\\cdot 4\\cdot 6\\cdot 7}{\\left(\\frac{1}{2}\\right)}^{7}+\\cdots \\right)\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nhas been attributed to Newton in the late [latex]1600\\text{s}\\text{.}[\/latex] The proof of this result uses the Maclaurin series for [latex]f\\left(x\\right)={\\sin}^{-1}x[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1169736851958\" type=\"a\">\n<li>Prove that the series converges.<\/li>\n<li>Evaluate the partial sums [latex]{S}_{n}[\/latex] for [latex]n=5,10,20[\/latex].<\/li>\n<li>Compare [latex]{S}_{n}[\/latex] to [latex]\\pi[\/latex] for [latex]n=5,10,20[\/latex] and discuss the number of correct decimal places.<\/li>\n<\/ol>\n<\/li>\n<li>The series<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738920677\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{\\pi }=\\frac{\\sqrt{8}}{9801}\\displaystyle\\sum _{n=0}^{\\infty }\\frac{\\left(4n\\right)\\text{!}\\left(1103+26390n\\right)}{{\\left(n\\text{!}\\right)}^{4}{396}^{4n}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwas discovered by <span class=\"no-emphasis\" data-type=\"term\">Ramanujan<\/span> in the early [latex]1900\\text{s}\\text{.}[\/latex] William Gosper, Jr., used this series to calculate [latex]\\pi[\/latex] to an accuracy of more than [latex]17[\/latex] million digits in the [latex]\\text{mid-}1980\\text{s}\\text{.}[\/latex] At the time, that was a world record. Since that time, this series and others by Ramanujan have led mathematicians to find many other series representations for [latex]\\pi[\/latex] and [latex]\\frac{1}{\\pi}[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<ol id=\"fs-id1169739252686\" type=\"a\">\n<li>Prove that this series converges.<\/li>\n<li>Evaluate the first term in this series. Compare this number with the value of [latex]\\pi[\/latex] from a calculating utility. To how many decimal places do these two numbers agree? What if we add the first two terms in the series?<\/li>\n<li>Investigate the life of Srinivasa Ramanujan [latex]\\left(1887\\text{-}1920\\right)[\/latex] and write a brief summary. Ramanujan is one of the most fascinating stories in the history of mathematics. He was basically self-taught, with no formal training in mathematics, yet he contributed in highly original ways to many advanced areas of mathematics.<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169739252731\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/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-1724\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Mixed Convergence Tests. <strong>Authored by<\/strong>: Ryan Melton. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/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":26,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"Mixed Convergence Tests\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"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-1724","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1724","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1724\/revisions"}],"predecessor-version":[{"id":2143,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1724\/revisions\/2143"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1724\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1724"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1724"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1724"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}