{"id":1714,"date":"2021-07-23T20:23:58","date_gmt":"2021-07-23T20:23:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1714"},"modified":"2022-03-21T23:05:17","modified_gmt":"2022-03-21T23:05:17","slug":"limit-comparison-test","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/limit-comparison-test\/","title":{"raw":"Limit Comparison Test","rendered":"Limit Comparison Test"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the limit comparison test to determine convergence of a series<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1169736613913\">The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series<\/p>\r\n\r\n<div id=\"fs-id1169736594105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739017630\">It is natural to compare this series with the convergent series<\/p>\r\n\r\n<div id=\"fs-id1169739017633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738986359\">However, this series does not satisfy the hypothesis necessary to use the comparison test because<\/p>\r\n\r\n<div id=\"fs-id1169738915172\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{n}^{2}-1}&gt;\\frac{1}{{n}^{2}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736766566\">for all integers [latex]n\\ge 2[\/latex]. Although we could look for a different series with which to compare [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\left({n}^{2}-1\\right)}[\/latex], instead we show how we can use the <strong>limit comparison test<\/strong> to compare<\/p>\r\n\r\n<div id=\"fs-id1169739269305\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}\\text{ and }\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739006239\">Let us examine the idea behind the limit comparison test. Consider two series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. with positive terms [latex]{a}_{n}\\text{and}{b}_{n}[\/latex] and evaluate<\/p>\r\n\r\n<div id=\"fs-id1169739030080\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739303327\">If<\/p>\r\n\r\n<div id=\"fs-id1169738869977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=L\\ne 0[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739109770\">then, for [latex]n[\/latex] sufficiently large, [latex]{a}_{n}\\approx L{b}_{n}[\/latex]. Therefore, either both series converge or both series diverge. For the series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\left({n}^{2}-1\\right)}[\/latex] and [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169736644288\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left({n}^{2}-1\\right)}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}}{{n}^{2}-1}=1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169736710484\">Since [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169739183274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738924055\">converges.<\/p>\r\n<p id=\"fs-id1169738962322\">The limit comparison test can be used in two other cases. Suppose<\/p>\r\n\r\n<div id=\"fs-id1169738962326\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739097066\">In this case, [latex]\\left\\{\\frac{{a}_{n}}{{b}_{n}}\\right\\}[\/latex] is a bounded sequence. As a result, there exists a constant [latex]M[\/latex] such that [latex]{a}_{n}\\le M{b}_{n}[\/latex]. Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. On the other hand, suppose<\/p>\r\n\r\n<div id=\"fs-id1169739187915\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738972186\">In this case,[latex]\\left\\{\\frac{{a}_{n}}{{b}_{n}}\\right\\}[\/latex] is an unbounded sequence. Therefore, for every constant [latex]M[\/latex] there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\ge M{b}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges as well.<\/p>\r\n\r\n<div id=\"fs-id1169739293675\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 data-type=\"title\">Theorem: Limit Comparison Test<\/h3>\r\n<p id=\"fs-id1169739099127\" style=\"text-align: left;\">Let [latex]{a}_{n},{b}_{n}\\ge 0[\/latex] for all [latex]n\\ge 1[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1169739029292\" type=\"i\">\r\n \t<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=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.<\/li>\r\n \t<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/li>\r\n \t<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty [\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169739297879\">Note that if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, the limit comparison test gives no information. Similarly, if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to \\infty [\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, the test also provides no information. For example, consider the two series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. These series are both <em data-effect=\"italics\">p<\/em>-series with [latex]p=\\frac{1}{2}[\/latex] and [latex]p=2[\/latex], respectively. Since [latex]p=\\frac{1}{2}&gt;1[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] diverges. On the other hand, since [latex]p=2&lt;1[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges. However, suppose we attempted to apply the limit comparison test, using the convergent [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] as our comparison series. First, we see that<\/p>\r\n\r\n<div id=\"fs-id1169738993576\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{\\sqrt{n}}}{\\frac{1}{{n}^{3}}}=\\frac{{n}^{3}}{\\sqrt{n}}={n}^{\\frac{5}{2}}\\to \\infty \\text{as}n\\to \\infty [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739215885\">Similarly, we see that<\/p>\r\n\r\n<div id=\"fs-id1169739215888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{{n}^{2}}}{\\frac{1}{{n}^{3}}}=n\\to \\infty \\text{as}n\\to \\infty [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169739304426\">Therefore, if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to \\infty [\/latex] when [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, we do not gain any information on the convergence or divergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738980859\" data-type=\"example\">\r\n<div id=\"fs-id1169738980862\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738980864\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Using the Limit Comparison Test<\/h3>\r\n<div id=\"fs-id1169738980864\" data-type=\"problem\">\r\n<p id=\"fs-id1169736728926\">For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.<\/p>\r\n\r\n<ol id=\"fs-id1169739040758\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1169738869108\" data-type=\"solution\">\r\n<ol id=\"fs-id1169738869110\" type=\"a\">\r\n \t<li>Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex]. Calculate<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738998893\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(\\sqrt{n}+1\\right)}}{\\frac{1}{\\sqrt{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\begin{array}{c}\\frac{\\sqrt{n}}{\\sqrt{n}+1}\\\\ \\end{array}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\sqrt{n}}}{1+\\frac{1}{\\sqrt{n}}}=1[\/latex].<\/div>\r\nBy the limit comparison test, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex] diverges.<\/li>\r\n \t<li>Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex]. We see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736613756\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{3}^{n}}\\cdot \\frac{{3}^{n}}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\left[1+{\\left(\\frac{1}{2}\\right)}^{n}\\right]=1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739222678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=1[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex] converges, we conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex] converges.<\/li>\r\n \t<li>Since [latex]\\text{ln}n&lt;n[\/latex], compare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. We see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739097242\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\underset{n\\to\\infty}\\lim} \\text{ln}\\frac{n}{n^2}\\frac{1}{n}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n^2}\\cdot \\frac{n}{1}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIn order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}[\/latex], evaluate the limit as [latex]x\\to \\infty [\/latex] of the real-valued function [latex]\\text{ln}\\frac{\\left(x\\right)}{x}[\/latex]. These two limits are equal, and making this change allows us to use L\u2019H\u00f4pital\u2019s rule. We obtain<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738904914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{1}{x}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}=0[\/latex], and, consequently,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169736654583\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{n}}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the limit is [latex]0[\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, the limit comparison test does not provide any information.<span data-type=\"newline\">\r\n<\/span>\r\nCompare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] instead. In this case,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739067731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{2}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}n=\\infty [\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the limit is [latex]\\infty [\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, the test still does not provide any information.<span data-type=\"newline\">\r\n<\/span>\r\nSo now we try a series between the two we already tried. Choosing the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex], we see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738869752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{\\frac{3}{2}}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{\\frac{3}{2}}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{\\sqrt{n}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nAs above, in order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{\\sqrt{n}}[\/latex], evaluate the limit as [latex]x\\to \\infty [\/latex] of the real-valued function [latex]\\text{ln}\\frac{x}{\\sqrt{x}}[\/latex]. Using L\u2019H\u00f4pital\u2019s rule,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169739210063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{\\sqrt{x}}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2\\sqrt{x}}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2}{\\sqrt{x}}=0[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nSince the limit is [latex]0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex] converges, we can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}n}{{n}^{2}}[\/latex] 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-id1169736587951\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169736587954\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738940786\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738940786\" data-type=\"problem\">\r\n<p id=\"fs-id1169738940789\">Use the limit comparison test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{5}^{n}}{{3}^{n}+2}[\/latex] converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169736778346\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169736778352\">Compare with a geometric series.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169739102660\" data-type=\"solution\">\r\n<p id=\"fs-id1169739102662\">The series diverges.<\/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\/vHBScF-UQZ4?controls=0&amp;start=555&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>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.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.4.2_555to664_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip \"5.4.2\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]230758[\/ohm_question]\r\n\r\n<\/div>\r\n<section id=\"fs-id1169736777480\" 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>Use the limit comparison test to determine convergence of a series<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1169736613913\">The comparison test works nicely if we can find a comparable series satisfying the hypothesis of the test. However, sometimes finding an appropriate series can be difficult. Consider the series<\/p>\n<div id=\"fs-id1169736594105\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739017630\">It is natural to compare this series with the convergent series<\/p>\n<div id=\"fs-id1169739017633\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738986359\">However, this series does not satisfy the hypothesis necessary to use the comparison test because<\/p>\n<div id=\"fs-id1169738915172\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{n}^{2}-1}>\\frac{1}{{n}^{2}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736766566\">for all integers [latex]n\\ge 2[\/latex]. Although we could look for a different series with which to compare [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\left({n}^{2}-1\\right)}[\/latex], instead we show how we can use the <strong>limit comparison test<\/strong> to compare<\/p>\n<div id=\"fs-id1169739269305\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}\\text{ and }\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739006239\">Let us examine the idea behind the limit comparison test. Consider two series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex]. with positive terms [latex]{a}_{n}\\text{and}{b}_{n}[\/latex] and evaluate<\/p>\n<div id=\"fs-id1169739030080\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739303327\">If<\/p>\n<div id=\"fs-id1169738869977\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=L\\ne 0[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739109770\">then, for [latex]n[\/latex] sufficiently large, [latex]{a}_{n}\\approx L{b}_{n}[\/latex]. Therefore, either both series converge or both series diverge. For the series [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{\\left({n}^{2}-1\\right)}[\/latex] and [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex], we see that<\/p>\n<div id=\"fs-id1169736644288\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left({n}^{2}-1\\right)}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{n}^{2}}{{n}^{2}-1}=1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169736710484\">Since [latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, we conclude that<\/p>\n<div id=\"fs-id1169739183274\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=2}^{\\infty }\\frac{1}{{n}^{2}-1}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738924055\">converges.<\/p>\n<p id=\"fs-id1169738962322\">The limit comparison test can be used in two other cases. Suppose<\/p>\n<div id=\"fs-id1169738962326\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739097066\">In this case, [latex]\\left\\{\\frac{{a}_{n}}{{b}_{n}}\\right\\}[\/latex] is a bounded sequence. As a result, there exists a constant [latex]M[\/latex] such that [latex]{a}_{n}\\le M{b}_{n}[\/latex]. Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges. On the other hand, suppose<\/p>\n<div id=\"fs-id1169739187915\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738972186\">In this case,[latex]\\left\\{\\frac{{a}_{n}}{{b}_{n}}\\right\\}[\/latex] is an unbounded sequence. Therefore, for every constant [latex]M[\/latex] there exists an integer [latex]N[\/latex] such that [latex]{a}_{n}\\ge M{b}_{n}[\/latex] for all [latex]n\\ge N[\/latex]. Therefore, if [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges as well.<\/p>\n<div id=\"fs-id1169739293675\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 data-type=\"title\">Theorem: Limit Comparison Test<\/h3>\n<p id=\"fs-id1169739099127\" style=\"text-align: left;\">Let [latex]{a}_{n},{b}_{n}\\ge 0[\/latex] for all [latex]n\\ge 1[\/latex].<\/p>\n<ol id=\"fs-id1169739029292\" type=\"i\">\n<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=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.<\/li>\n<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges.<\/li>\n<li>If [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{{a}_{n}}{{b}_{n}}=\\infty[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169739297879\">Note that if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to 0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] diverges, the limit comparison test gives no information. Similarly, if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to \\infty[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, the test also provides no information. For example, consider the two series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex]. These series are both <em data-effect=\"italics\">p<\/em>-series with [latex]p=\\frac{1}{2}[\/latex] and [latex]p=2[\/latex], respectively. Since [latex]p=\\frac{1}{2}>1[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] diverges. On the other hand, since [latex]p=2<1[\/latex], the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges. However, suppose we attempted to apply the limit comparison test, using the convergent [latex]p-\\text{series}[\/latex] [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] as our comparison series. First, we see that<\/p>\n<div id=\"fs-id1169738993576\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{\\sqrt{n}}}{\\frac{1}{{n}^{3}}}=\\frac{{n}^{3}}{\\sqrt{n}}={n}^{\\frac{5}{2}}\\to \\infty \\text{as}n\\to \\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739215885\">Similarly, we see that<\/p>\n<div id=\"fs-id1169739215888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{\\frac{1}{{n}^{2}}}{\\frac{1}{{n}^{3}}}=n\\to \\infty \\text{as}n\\to \\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169739304426\">Therefore, if [latex]\\frac{{a}_{n}}{{b}_{n}}\\to \\infty[\/latex] when [latex]\\displaystyle\\sum _{n=1}^{\\infty }{b}_{n}[\/latex] converges, we do not gain any information on the convergence or divergence of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex].<\/p>\n<div id=\"fs-id1169738980859\" data-type=\"example\">\n<div id=\"fs-id1169738980862\" data-type=\"exercise\">\n<div id=\"fs-id1169738980864\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Using the Limit Comparison Test<\/h3>\n<div id=\"fs-id1169738980864\" data-type=\"problem\">\n<p id=\"fs-id1169736728926\">For each of the following series, use the limit comparison test to determine whether the series converges or diverges. If the test does not apply, say so.<\/p>\n<ol id=\"fs-id1169739040758\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}\\left(n\\right)}{{n}^{2}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738869108\" data-type=\"solution\">\n<ol id=\"fs-id1169738869110\" type=\"a\">\n<li>Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex]. Calculate<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738998893\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\left(\\sqrt{n}+1\\right)}}{\\frac{1}{\\sqrt{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\begin{array}{c}\\frac{\\sqrt{n}}{\\sqrt{n}+1}\\\\ \\end{array}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{1}{\\sqrt{n}}}{1+\\frac{1}{\\sqrt{n}}}=1[\/latex].<\/div>\n<p>By the limit comparison test, since [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}}[\/latex] diverges, then [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{\\sqrt{n}+1}[\/latex] diverges.<\/li>\n<li>Compare this series to [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex]. We see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736613756\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{3}^{n}}\\cdot \\frac{{3}^{n}}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{{2}^{n}+1}{{2}^{n}}=\\underset{n\\to \\infty }{\\text{lim}}\\left[1+{\\left(\\frac{1}{2}\\right)}^{n}\\right]=1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739222678\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\frac{\\left({2}^{n}+1\\right)}{{3}^{n}}}{\\frac{{2}^{n}}{{3}^{n}}}=1[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]\\displaystyle\\sum _{n=1}^{\\infty }{\\left(\\frac{2}{3}\\right)}^{n}[\/latex] converges, we conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{2}^{n}+1}{{3}^{n}}[\/latex] converges.<\/li>\n<li>Since [latex]\\text{ln}n<n[\/latex], compare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex]. We see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739097242\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\underset{n\\to\\infty}\\lim} \\text{ln}\\frac{n}{n^2}\\frac{1}{n}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n^2}\\cdot \\frac{n}{1}= {\\underset{n\\to\\infty}\\lim} \\frac{\\text{ln}n}{n}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}[\/latex], evaluate the limit as [latex]x\\to \\infty[\/latex] of the real-valued function [latex]\\text{ln}\\frac{\\left(x\\right)}{x}[\/latex]. These two limits are equal, and making this change allows us to use L\u2019H\u00f4pital\u2019s rule. We obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738904914\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{1}{x}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{n}=0[\/latex], and, consequently,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169736654583\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{n}}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the limit is [latex]0[\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges, the limit comparison test does not provide any information.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nCompare with [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] instead. In this case,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739067731\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{2}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{2}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}n=\\infty[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the limit is [latex]\\infty[\/latex] but [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] converges, the test still does not provide any information.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSo now we try a series between the two we already tried. Choosing the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex], we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738869752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}\\frac{n}{{n}^{2}}}{\\frac{1}{{n}^{\\frac{3}{2}}}}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{{n}^{2}}\\cdot \\frac{{n}^{\\frac{3}{2}}}{1}=\\underset{n\\to \\infty }{\\text{lim}}\\frac{\\text{ln}n}{\\sqrt{n}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAs above, in order to evaluate [latex]\\underset{n\\to \\infty }{\\text{lim}}\\text{ln}\\frac{n}{\\sqrt{n}}[\/latex], evaluate the limit as [latex]x\\to \\infty[\/latex] of the real-valued function [latex]\\text{ln}\\frac{x}{\\sqrt{x}}[\/latex]. Using L\u2019H\u00f4pital\u2019s rule,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169739210063\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\underset{x\\to \\infty }{\\text{lim}}\\frac{\\text{ln}x}{\\sqrt{x}}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2\\sqrt{x}}{x}=\\underset{x\\to \\infty }{\\text{lim}}\\frac{2}{\\sqrt{x}}=0[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince the limit is [latex]0[\/latex] and [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{3}{2}}}[\/latex] converges, we can conclude that [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{\\text{ln}n}{{n}^{2}}[\/latex] converges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169736587951\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169736587954\" data-type=\"exercise\">\n<div id=\"fs-id1169738940786\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738940786\" data-type=\"problem\">\n<p id=\"fs-id1169738940789\">Use the limit comparison test to determine whether the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{{5}^{n}}{{3}^{n}+2}[\/latex] converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736778346\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169736778352\">Compare with a geometric series.<\/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=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739102660\" data-type=\"solution\">\n<p id=\"fs-id1169739102662\">The series diverges.<\/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\/vHBScF-UQZ4?controls=0&amp;start=555&amp;end=664&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>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\/5.4.2_555to664_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip &#8220;5.4.2&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm230758\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=230758&theme=oea&iframe_resize_id=ohm230758&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<section id=\"fs-id1169736777480\" 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-1714\">\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>5.4.2. <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":18,"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\":\"5.4.2\",\"author\":\"\",\"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-1714","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1714","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\/1714\/revisions"}],"predecessor-version":[{"id":2218,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1714\/revisions\/2218"}],"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\/1714\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1714"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1714"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1714"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1714"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}