{"id":1770,"date":"2021-07-26T22:19:15","date_gmt":"2021-07-26T22:19:15","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1770"},"modified":"2022-03-21T23:04:27","modified_gmt":"2022-03-21T23:04:27","slug":"the-p-series-and-estimating-series-value","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/the-p-series-and-estimating-series-value\/","title":{"raw":"The p-Series and Estimating Series Value","rendered":"The p-Series and Estimating Series Value"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Estimate the value of a series by finding bounds on its remainder term<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">The <em data-effect=\"italics\">p<\/em>-Series<\/h2>\r\n<p id=\"fs-id1169738055544\">The harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] are both examples of a type of series called a <em data-effect=\"italics\">p<\/em>-series.<\/p>\r\n\r\n<div id=\"fs-id1169737809245\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737930774\">For any real number [latex]p[\/latex], the series<\/p>\r\n\r\n<div id=\"fs-id1169737232687\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737742410\">is called a <strong><em data-effect=\"italics\">p<\/em>-series<\/strong>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169737287287\">We know the <em data-effect=\"italics\">p<\/em>-series converges if [latex]p=2[\/latex] and diverges if [latex]p=1[\/latex]. What about other values of [latex]p\\text{?}[\/latex] In general, it is difficult, if not impossible, to compute the exact value of most [latex]p[\/latex] -series. However, we can use the tests presented thus far to prove whether a [latex]p[\/latex] -series converges or diverges.<\/p>\r\n<p id=\"fs-id1169737759522\" style=\"text-align: left;\">If [latex]p&lt;0[\/latex], then [latex]\\frac{1}{{n}^{p}}\\to \\infty [\/latex], and if [latex]p=0[\/latex], then [latex]\\frac{1}{{n}^{p}}\\to 1[\/latex]. Therefore, by the divergence test,<\/p>\r\n\r\n<div id=\"fs-id1169738249556\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}\\text{ diverges if }p\\le 0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738144952\">If [latex]p&gt;0[\/latex], then [latex]f\\left(x\\right)=\\frac{1}{{x}^{p}}[\/latex] is a positive, continuous, decreasing function. Therefore, for [latex]p&gt;0[\/latex], we use the integral test, comparing<\/p>\r\n\r\n<div id=\"fs-id1169737796258\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}\\text{ and }{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737161035\">We have already considered the case when [latex]p=1[\/latex]. Here we consider the case when [latex]p&gt;0,p\\ne 1[\/latex]. For this case,<\/p>\r\n\r\n<div id=\"fs-id1169737185451\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}\\left[{b}^{1-p}-1\\right][\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737434844\">Because<\/p>\r\n\r\n<div id=\"fs-id1169738180046\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{b}^{1-p}\\to 0\\text{ if }p&gt;1\\text{ and }{b}^{1-p}\\to \\infty \\text{ if }p&lt;1[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738143759\">we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738143762\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int _{1}^{\\infty}} \\dfrac{1}{x^{p}}dx = \\Bigg\\{ \\begin{array}{c} \\frac{1}{p-1}\\text{ if }p&gt;1\\\\ \\infty \\text{ if }p&lt;1\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737430026\">Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] converges if [latex]p&gt;1[\/latex] and diverges if [latex]0&lt;p&lt;1[\/latex].<\/p>\r\n<p id=\"fs-id1169738111508\">In summary,<\/p>\r\n\r\n<div id=\"fs-id1169738111511\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\sum _{n=1}^{\\infty}} \\dfrac{1}{n^{p}} \\bigg\\{ \\begin{array}{l} \\text{ converges if }p&gt;1\\\\ \\text{ diverges if }p\\le 1\\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<div id=\"fs-id1169737151882\" data-type=\"example\">\r\n<div id=\"fs-id1169737151884\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737151886\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Testing for Convergence of <em data-effect=\"italics\">p<\/em>-series<\/h3>\r\n<div id=\"fs-id1169737151886\" data-type=\"problem\">\r\n<p id=\"fs-id1169738143748\">For each of the following series, determine whether it converges or diverges.<\/p>\r\n\r\n<ol id=\"fs-id1169738143751\" type=\"a\">\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex]<\/li>\r\n \t<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{2}{3}}}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558884\"]\r\n<div id=\"fs-id1169737934869\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737934871\" type=\"a\">\r\n \t<li>This is a <em data-effect=\"italics\">p<\/em>-series with [latex]p=4&gt;1[\/latex], so the series converges.<\/li>\r\n \t<li>Since [latex]p=\\frac{2}{3}&lt;1[\/latex], the series diverges.<\/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-id1169737433536\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738055554\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\r\n<p id=\"fs-id1169738055558\">Does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{5}{4}}}[\/latex] converge or diverge?<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1169737209098\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169738080287\">[latex]p=\\frac{5}{4}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1169737209091\" data-type=\"solution\">\r\n<p id=\"fs-id1169737209093\">The series converges.<\/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\/1HTIl9x92UU?controls=0&amp;start=32&amp;end=44&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\/5.3.4_32to44_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.3.4\" here (opens in new window)<\/a>.\r\n\r\n<section id=\"fs-id1169737141433\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Estimating the Value of a Series<\/h2>\r\n<p id=\"fs-id1169737141439\">Suppose we know that a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] where [latex]N[\/latex] is any positive integer. The question we address here is, for a convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], how good is the approximation [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}\\text{?}[\/latex] More specifically, if we let<\/p>\r\n\r\n<div id=\"fs-id1169737933561\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737292618\">be the remainder when the sum of an infinite series is approximated by the [latex]N\\text{th}[\/latex] partial sum, how large is [latex]{R}_{N}\\text{?}[\/latex] For some types of series, we are able to use the ideas from the integral test to estimate [latex]{R}_{N}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169737162226\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Remainder Estimate from the Integral Test<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1169737394062\">Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a convergent series with positive terms. Suppose there exists a function [latex]f[\/latex] satisfying the following three conditions:<\/p>\r\n\r\n<ol id=\"fs-id1169737272110\" type=\"i\">\r\n \t<li>[latex]f[\/latex] is continuous,<\/li>\r\n \t<li>[latex]f[\/latex] is decreasing, and<\/li>\r\n \t<li>[latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge 1[\/latex].<\/li>\r\n<\/ol>\r\n<p id=\"fs-id1169738078014\">Let [latex]{S}_{N}[\/latex] be the <em data-effect=\"italics\">N<\/em>th partial sum of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. For all positive integers [latex]N[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1169737162233\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx&lt;\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}&lt;{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737299918\">In other words, the remainder [latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-{S}_{N}=\\displaystyle\\sum _{n=N+1}^{\\infty }{a}_{n}[\/latex] satisfies the following estimate:<\/p>\r\n\r\n<div id=\"fs-id1169738183813\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx&lt;{R}_{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737931127\">This is known as the <span data-type=\"term\">remainder estimate<\/span>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169737233552\">We illustrate the Remainder Estimate from the Integral Test in Figure 4. In particular, by representing the remainder [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots [\/latex] as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] and bounded below by [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. In other words,<\/p>\r\n\r\n<div id=\"fs-id1169738198566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &gt;{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738125850\">and<\/p>\r\n\r\n<div id=\"fs-id1169738125853\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &lt;{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738063663\">We conclude that<\/p>\r\n\r\n<div id=\"fs-id1169737301598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx&lt;{R}_{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737359282\">Since<\/p>\r\n\r\n<div id=\"fs-id1169737359285\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={S}_{N}+{R}_{N}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737429946\">where [latex]{S}_{N}[\/latex] is the [latex]N\\text{th}[\/latex] partial sum, we conclude that<\/p>\r\n\r\n<div id=\"fs-id1169738178546\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx&lt;\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}&lt;{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_09_03_004\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234351\/CNX_Calc_Figure_09_03_003.jpg\" alt=\"This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x-axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).\" width=\"731\" height=\"313\" data-media-type=\"image\/jpeg\" \/> Figure 4. Given a continuous, positive, decreasing function [latex]f[\/latex] and a sequence of positive terms [latex]{a}_{n}[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all positive integers [latex]n[\/latex], (a) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &lt;{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex], or (b) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &gt;{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. Therefore, the integral is either an overestimate or an underestimate of the error.[\/caption]<\/figure>\r\n<div id=\"fs-id1169737360049\" data-type=\"example\">\r\n<div id=\"fs-id1169737360052\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737360054\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Estimating the Value of a Series<\/h3>\r\n<div id=\"fs-id1169737360054\" data-type=\"problem\">\r\n<p id=\"fs-id1169737360059\">Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1169738193755\" type=\"a\">\r\n \t<li>Calculate [latex]{S}_{10}=\\displaystyle\\sum _{n=1}^{10}\\frac{1}{{n}^{3}}[\/latex] and estimate the error.<\/li>\r\n \t<li>Determine the least value of [latex]N[\/latex] necessary such that [latex]{S}_{N}[\/latex] will estimate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] to within [latex]0.001[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558882\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558882\"]\r\n<div id=\"fs-id1169737848229\" data-type=\"solution\">\r\n<ol id=\"fs-id1169737848231\" type=\"a\">\r\n \t<li>Using a calculating utility, we have<span data-type=\"newline\">\r\n<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div id=\"fs-id1169737848242\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{S}_{10}=1+\\frac{1}{{2}^{3}}+\\frac{1}{{3}^{3}}+\\frac{1}{{4}^{3}}+\\cdots +\\frac{1}{{10}^{3}}\\approx 1.19753[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nBy the remainder estimate, we know<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737361700\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nWe have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737392470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int _{10}^{\\infty }}\\frac{1}{{x}^{3}}dx= {\\underset{b\\to \\infty }\\lim} {\\displaystyle\\int _{10}^{b}} \\frac{1}{{x}^{3}}dx= {\\underset{b\\to\\infty}\\lim} \\left[ -\\frac{1}{2x^2} \\right] _{N}^{b}= {\\underset{b\\to \\infty }\\lim} \\left[-\\frac{1}{2{b}^{2}} + \\frac{1}{2{N}^{2}}\\right]= \\frac{1}{2{N}^{2}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore, the error is [latex]{R}_{10}&lt;\\frac{1}{2{\\left(10\\right)}^{2}}=0.005[\/latex].\r\n<ul>\r\n \t<li>Find [latex]N[\/latex] such that [latex]{R}_{N}&lt;0.001[\/latex]. In part a. we showed that [latex]{R}_{N}&lt;\\frac{1}{2{N}^{2}}[\/latex]. Therefore, the remainder [latex]{R}_{N}&lt;0.001[\/latex] as long as [latex]\\frac{1}{2{N}^{2}}&lt;0.001[\/latex]. That is, we need [latex]2{N}^{2}&gt;1000[\/latex]. Solving this inequality for [latex]N[\/latex], we see that we need [latex]N&gt;22.36[\/latex]. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is [latex]N=23[\/latex].<\/li>\r\n<\/ul>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n<div id=\"fs-id1169738249580\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738249584\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\r\n<p id=\"fs-id1169738249588\">For [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex], calculate [latex]{S}_{5}[\/latex] and estimate the error [latex]{R}_{5}[\/latex].<\/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-id1169737265746\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737265752\">Use the remainder estimate [latex]{R}_{N}&lt;{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{4}}dx[\/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-id1169737394110\" data-type=\"solution\">\r\n<p id=\"fs-id1169737394112\">[latex]{S}_{5}\\approx 1.09035[\/latex], [latex]{R}_{5}&lt;0.00267[\/latex]<\/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\n<section id=\"fs-id1169738052811\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\">\r\n<dl id=\"fs-id1169737174578\">\r\n \t<dd id=\"fs-id1169737174583\">\r\n<div id=\"fs-id1169738066622\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div><\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Estimate the value of a series by finding bounds on its remainder term<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">The <em data-effect=\"italics\">p<\/em>-Series<\/h2>\n<p id=\"fs-id1169738055544\">The harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] and the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{2}}[\/latex] are both examples of a type of series called a <em data-effect=\"italics\">p<\/em>-series.<\/p>\n<div id=\"fs-id1169737809245\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1169737930774\">For any real number [latex]p[\/latex], the series<\/p>\n<div id=\"fs-id1169737232687\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737742410\">is called a <strong><em data-effect=\"italics\">p<\/em>-series<\/strong>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737287287\">We know the <em data-effect=\"italics\">p<\/em>-series converges if [latex]p=2[\/latex] and diverges if [latex]p=1[\/latex]. What about other values of [latex]p\\text{?}[\/latex] In general, it is difficult, if not impossible, to compute the exact value of most [latex]p[\/latex] -series. However, we can use the tests presented thus far to prove whether a [latex]p[\/latex] -series converges or diverges.<\/p>\n<p id=\"fs-id1169737759522\" style=\"text-align: left;\">If [latex]p<0[\/latex], then [latex]\\frac{1}{{n}^{p}}\\to \\infty[\/latex], and if [latex]p=0[\/latex], then [latex]\\frac{1}{{n}^{p}}\\to 1[\/latex]. Therefore, by the divergence test,<\/p>\n<div id=\"fs-id1169738249556\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}\\text{ diverges if }p\\le 0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738144952\">If [latex]p>0[\/latex], then [latex]f\\left(x\\right)=\\frac{1}{{x}^{p}}[\/latex] is a positive, continuous, decreasing function. Therefore, for [latex]p>0[\/latex], we use the integral test, comparing<\/p>\n<div id=\"fs-id1169737796258\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}\\text{ and }{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737161035\">We have already considered the case when [latex]p=1[\/latex]. Here we consider the case when [latex]p>0,p\\ne 1[\/latex]. For this case,<\/p>\n<div id=\"fs-id1169737185451\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{b}\\frac{1}{{x}^{p}}dx=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}{x}^{1-p}{|}_{1}^{b}=\\underset{b\\to \\infty }{\\text{lim}}\\frac{1}{1-p}\\left[{b}^{1-p}-1\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737434844\">Because<\/p>\n<div id=\"fs-id1169738180046\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{b}^{1-p}\\to 0\\text{ if }p>1\\text{ and }{b}^{1-p}\\to \\infty \\text{ if }p<1[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738143759\">we conclude that<\/p>\n<div id=\"fs-id1169738143762\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int _{1}^{\\infty}} \\dfrac{1}{x^{p}}dx = \\Bigg\\{ \\begin{array}{c} \\frac{1}{p-1}\\text{ if }p>1\\\\ \\infty \\text{ if }p<1\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737430026\">Therefore, [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{p}}[\/latex] converges if [latex]p>1[\/latex] and diverges if [latex]0<p<1[\/latex].<\/p>\n<p id=\"fs-id1169738111508\">In summary,<\/p>\n<div id=\"fs-id1169738111511\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\sum _{n=1}^{\\infty}} \\dfrac{1}{n^{p}} \\bigg\\{ \\begin{array}{l} \\text{ converges if }p>1\\\\ \\text{ diverges if }p\\le 1\\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1169737151882\" data-type=\"example\">\n<div id=\"fs-id1169737151884\" data-type=\"exercise\">\n<div id=\"fs-id1169737151886\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Testing for Convergence of <em data-effect=\"italics\">p<\/em>-series<\/h3>\n<div id=\"fs-id1169737151886\" data-type=\"problem\">\n<p id=\"fs-id1169738143748\">For each of the following series, determine whether it converges or diverges.<\/p>\n<ol id=\"fs-id1169738143751\" type=\"a\">\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex]<\/li>\n<li>[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{2}{3}}}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558884\">Show Solution<\/span><\/p>\n<div id=\"q44558884\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737934869\" data-type=\"solution\">\n<ol id=\"fs-id1169737934871\" type=\"a\">\n<li>This is a <em data-effect=\"italics\">p<\/em>-series with [latex]p=4>1[\/latex], so the series converges.<\/li>\n<li>Since [latex]p=\\frac{2}{3}<1[\/latex], the series diverges.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169737433536\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738055554\" data-type=\"exercise\">\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738055556\" data-type=\"problem\">\n<p id=\"fs-id1169738055558\">Does the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{\\frac{5}{4}}}[\/latex] converge or diverge?<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Hint<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737209098\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169738080287\">[latex]p=\\frac{5}{4}[\/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=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737209091\" data-type=\"solution\">\n<p id=\"fs-id1169737209093\">The series converges.<\/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\/1HTIl9x92UU?controls=0&amp;start=32&amp;end=44&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\/5.3.4_32to44_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.3.4&#8221; here (opens in new window)<\/a>.<\/p>\n<section id=\"fs-id1169737141433\" data-depth=\"1\">\n<h2 data-type=\"title\">Estimating the Value of a Series<\/h2>\n<p id=\"fs-id1169737141439\">Suppose we know that a series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] converges and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex] where [latex]N[\/latex] is any positive integer. The question we address here is, for a convergent series [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex], how good is the approximation [latex]\\displaystyle\\sum _{n=1}^{N}{a}_{n}\\text{?}[\/latex] More specifically, if we let<\/p>\n<div id=\"fs-id1169737933561\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-\\displaystyle\\sum _{n=1}^{N}{a}_{n}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737292618\">be the remainder when the sum of an infinite series is approximated by the [latex]N\\text{th}[\/latex] partial sum, how large is [latex]{R}_{N}\\text{?}[\/latex] For some types of series, we are able to use the ideas from the integral test to estimate [latex]{R}_{N}[\/latex].<\/p>\n<div id=\"fs-id1169737162226\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">theorem: Remainder Estimate from the Integral Test<\/h3>\n<hr \/>\n<p id=\"fs-id1169737394062\">Suppose [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex] is a convergent series with positive terms. Suppose there exists a function [latex]f[\/latex] satisfying the following three conditions:<\/p>\n<ol id=\"fs-id1169737272110\" type=\"i\">\n<li>[latex]f[\/latex] is continuous,<\/li>\n<li>[latex]f[\/latex] is decreasing, and<\/li>\n<li>[latex]f\\left(n\\right)={a}_{n}[\/latex] for all integers [latex]n\\ge 1[\/latex].<\/li>\n<\/ol>\n<p id=\"fs-id1169738078014\">Let [latex]{S}_{N}[\/latex] be the <em data-effect=\"italics\">N<\/em>th partial sum of [latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}[\/latex]. For all positive integers [latex]N[\/latex],<\/p>\n<div id=\"fs-id1169737162233\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737299918\">In other words, the remainder [latex]{R}_{N}=\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}-{S}_{N}=\\displaystyle\\sum _{n=N+1}^{\\infty }{a}_{n}[\/latex] satisfies the following estimate:<\/p>\n<div id=\"fs-id1169738183813\" style=\"text-align: center;\" data-type=\"equation\">[latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737931127\">This is known as the <span data-type=\"term\">remainder estimate<\/span>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169737233552\">We illustrate the Remainder Estimate from the Integral Test in Figure 4. In particular, by representing the remainder [latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots[\/latex] as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by [latex]{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex] and bounded below by [latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. In other words,<\/p>\n<div id=\"fs-id1169738198566\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots >{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738125850\">and<\/p>\n<div id=\"fs-id1169738125853\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}={a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots <{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738063663\">We conclude that<\/p>\n<div id=\"fs-id1169737301598\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737359282\">Since<\/p>\n<div id=\"fs-id1169737359285\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}={S}_{N}+{R}_{N}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737429946\">where [latex]{S}_{N}[\/latex] is the [latex]N\\text{th}[\/latex] partial sum, we conclude that<\/p>\n<div id=\"fs-id1169738178546\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{N}+{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx<\\displaystyle\\sum _{n=1}^{\\infty }{a}_{n}<{S}_{N}+{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_09_03_004\"><figcaption><\/figcaption><div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234351\/CNX_Calc_Figure_09_03_003.jpg\" alt=\"This shows two graphs side by side of the same decreasing concave up function y = f(x) that approaches the x-axis in quadrant 1. Rectangles are drawn with a base of 1 over the intervals N through N + 4. The heights of the rectangles in the first graph are determined by the value of the function at the right endpoints of the bases, and those in the second graph are determined by the value at the left endpoints of the bases. The areas of the rectangles are marked: a_(N + 1), a_(N + 2), through a_(N + 4).\" width=\"731\" height=\"313\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Given a continuous, positive, decreasing function [latex]f[\/latex] and a sequence of positive terms [latex]{a}_{n}[\/latex] such that [latex]{a}_{n}=f\\left(n\\right)[\/latex] for all positive integers [latex]n[\/latex], (a) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &lt;{\\displaystyle\\int }_{N}^{\\infty }f\\left(x\\right)dx[\/latex], or (b) the areas [latex]{a}_{N+1}+{a}_{N+2}+{a}_{N+3}+\\cdots &gt;{\\displaystyle\\int }_{N+1}^{\\infty }f\\left(x\\right)dx[\/latex]. Therefore, the integral is either an overestimate or an underestimate of the error.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1169737360049\" data-type=\"example\">\n<div id=\"fs-id1169737360052\" data-type=\"exercise\">\n<div id=\"fs-id1169737360054\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Estimating the Value of a Series<\/h3>\n<div id=\"fs-id1169737360054\" data-type=\"problem\">\n<p id=\"fs-id1169737360059\">Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex].<\/p>\n<ol id=\"fs-id1169738193755\" type=\"a\">\n<li>Calculate [latex]{S}_{10}=\\displaystyle\\sum _{n=1}^{10}\\frac{1}{{n}^{3}}[\/latex] and estimate the error.<\/li>\n<li>Determine the least value of [latex]N[\/latex] necessary such that [latex]{S}_{N}[\/latex] will estimate [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{3}}[\/latex] to within [latex]0.001[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558882\">Show Solution<\/span><\/p>\n<div id=\"q44558882\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737848229\" data-type=\"solution\">\n<ol id=\"fs-id1169737848231\" type=\"a\">\n<li>Using a calculating utility, we have<span data-type=\"newline\"><br \/>\n<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div id=\"fs-id1169737848242\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]{S}_{10}=1+\\frac{1}{{2}^{3}}+\\frac{1}{{3}^{3}}+\\frac{1}{{4}^{3}}+\\cdots +\\frac{1}{{10}^{3}}\\approx 1.19753[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nBy the remainder estimate, we know<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737361700\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{3}}dx[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nWe have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737392470\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int _{10}^{\\infty }}\\frac{1}{{x}^{3}}dx= {\\underset{b\\to \\infty }\\lim} {\\displaystyle\\int _{10}^{b}} \\frac{1}{{x}^{3}}dx= {\\underset{b\\to\\infty}\\lim} \\left[ -\\frac{1}{2x^2} \\right] _{N}^{b}= {\\underset{b\\to \\infty }\\lim} \\left[-\\frac{1}{2{b}^{2}} + \\frac{1}{2{N}^{2}}\\right]= \\frac{1}{2{N}^{2}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore, the error is [latex]{R}_{10}<\\frac{1}{2{\\left(10\\right)}^{2}}=0.005[\/latex].\n\n\n<ul>\n<li>Find [latex]N[\/latex] such that [latex]{R}_{N}<0.001[\/latex]. In part a. we showed that [latex]{R}_{N}<\\frac{1}{2{N}^{2}}[\/latex]. Therefore, the remainder [latex]{R}_{N}<0.001[\/latex] as long as [latex]\\frac{1}{2{N}^{2}}<0.001[\/latex]. That is, we need [latex]2{N}^{2}>1000[\/latex]. Solving this inequality for [latex]N[\/latex], we see that we need [latex]N>22.36[\/latex]. To ensure that the remainder is within the desired amount, we need to round up to the nearest integer. Therefore, the minimum necessary value is [latex]N=23[\/latex].<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738249580\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738249584\" data-type=\"exercise\">\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738249586\" data-type=\"problem\">\n<p id=\"fs-id1169738249588\">For [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{{n}^{4}}[\/latex], calculate [latex]{S}_{5}[\/latex] and estimate the error [latex]{R}_{5}[\/latex].<\/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-id1169737265746\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737265752\">Use the remainder estimate [latex]{R}_{N}<{\\displaystyle\\int }_{N}^{\\infty }\\frac{1}{{x}^{4}}dx[\/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-id1169737394110\" data-type=\"solution\">\n<p id=\"fs-id1169737394112\">[latex]{S}_{5}\\approx 1.09035[\/latex], [latex]{R}_{5}<0.00267[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169738052811\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\">\n<dl id=\"fs-id1169737174578\">\n<dd id=\"fs-id1169737174583\">\n<div id=\"fs-id1169738066622\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<\/dd>\n<\/dl>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1770\">\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>5.3.4. <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":14,"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\":\"5.3.4\",\"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-1770","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1770","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\/1770\/revisions"}],"predecessor-version":[{"id":2668,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1770\/revisions\/2668"}],"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\/1770\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1770"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1770"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1770"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1770"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}