{"id":1700,"date":"2021-07-23T19:30:42","date_gmt":"2021-07-23T19:30:42","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1700"},"modified":"2022-03-21T23:03:27","modified_gmt":"2022-03-21T23:03:27","slug":"telescoping-series","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/telescoping-series\/","title":{"raw":"Telescoping Series","rendered":"Telescoping Series"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Evaluate a telescoping series<\/li>\r\n<\/ul>\r\n<\/div>\r\nConsider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex]. We discussed this series in the example: Evaluating Limits of Sequences of Partial Sums, showing that the series converges by writing out the first several partial sums [latex]{S}_{1},{S}_{2}\\text{,}\\ldots,{S}_{6}[\/latex] and noticing that they are all of the form [latex]{S}_{k}=\\frac{k}{k+1}[\/latex]. Here we use a different technique to show that this series converges. By using partial fractions, we can write\r\n<div id=\"fs-id1169738154896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{n\\left(n+1\\right)}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738216460\">Therefore, the series can be written as<\/p>\r\n\r\n<div id=\"fs-id1169738216464\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{1}{n}-\\frac{1}{n+1}\\right]=\\left(1+\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738191325\">Writing out the first several terms in the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1169738191343\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=1-\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=1-\\frac{1}{3}\\hfill \\\\ {S}_{3}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)=1-\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737910267\">In general,<\/p>\r\n\r\n<div id=\"fs-id1169737910270\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{k+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169738115213\">We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since [latex]{S}_{k}=1 - \\frac{1}{\\left(k+1\\right)}[\/latex] and [latex]\\frac{1}{\\left(k+1\\right)}\\to 0[\/latex] as [latex]k\\to \\infty [\/latex], the sequence of partial sums converges to [latex]1[\/latex], and therefore the series converges to [latex]1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1169738077699\" 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-id1169738077703\">A <strong>telescoping series<\/strong> is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1169738077712\">For example, any series of the form<\/p>\r\n\r\n<div id=\"fs-id1169738077715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)+\\cdots [\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737394662\">is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that<\/p>\r\n\r\n<div id=\"fs-id1169737394667\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\\hfill \\\\ {S}_{2}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)={b}_{1}-{b}_{3}\\hfill \\\\ {S}_{3}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)={b}_{1}-{b}_{4}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737168045\">In general, the <em data-effect=\"italics\">k<\/em>th partial sum of this series is<\/p>\r\n\r\n<div id=\"fs-id1169737168054\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={b}_{1}-{b}_{k+1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737168089\">Since the <em data-effect=\"italics\">k<\/em>th partial sum can be simplified to the difference of these two terms, the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] will converge if and only if the sequence [latex]\\left\\{{b}_{k+1}\\right\\}[\/latex] converges. Moreover, if the sequence [latex]{b}_{k+1}[\/latex] converges to some finite number [latex]B[\/latex], then the sequence of partial sums converges to [latex]{b}_{1}-B[\/latex], and therefore<\/p>\r\n\r\n<div id=\"fs-id1169738180671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]={b}_{1}-B[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737264635\">In the next example, we show how to use these ideas to analyze a telescoping series of this form.<\/p>\r\n\r\n<div id=\"fs-id1169737264640\" data-type=\"example\">\r\n<div id=\"fs-id1169737264642\" data-type=\"exercise\">\r\n<div id=\"fs-id1169737264644\" data-type=\"problem\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Evaluating a Telescoping Series<\/h3>\r\n<div id=\"fs-id1169737264644\" data-type=\"problem\">\r\n<p id=\"fs-id1169737264649\">Determine whether the telescoping series<\/p>\r\n\r\n<div id=\"fs-id1169737264652\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right][\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737154216\">converges or diverges. If it converges, find its sum.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1169737154220\" data-type=\"solution\">\r\n<p id=\"fs-id1169737154222\">By writing out terms in the sequence of partial sums, we can see that<\/p>\r\n\r\n<div id=\"fs-id1169737154225\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {S}_{1}&amp; =\\hfill &amp; \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\hfill \\\\ \\hfill {S}_{2}&amp; =\\hfill &amp; \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{3}\\right)\\hfill \\\\ \\hfill {S}_{3}&amp; =\\hfill &amp; \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{3}\\right)-\\cos\\left(\\frac{1}{4}\\right)\\right)\\hfill \\\\ &amp; =\\hfill &amp; \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{4}\\right).\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737392321\">In general,<\/p>\r\n\r\n<div id=\"fs-id1169737392324\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{k+1}\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737392372\">Since [latex]\\frac{1}{\\left(k+1\\right)}\\to 0[\/latex] as [latex]k\\to \\infty [\/latex] and [latex]\\cos{x}[\/latex] is a continuous function, [latex]\\cos\\left(\\frac{1}{\\left(k+1\\right)}\\right)\\to \\cos\\left(0\\right)=1[\/latex]. Therefore, we conclude that [latex]{S}_{k}\\to \\cos\\left(1\\right)-1[\/latex]. The telescoping series converges and the sum is given by<\/p>\r\n\r\n<div id=\"fs-id1169738185032\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right]=\\cos\\left(1\\right)-1[\/latex].<\/div>\r\n&nbsp;\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<div id=\"fs-id1169738185120\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1169738185124\" data-type=\"exercise\">\r\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\r\n<p id=\"fs-id1169738185128\">Determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{e}^{\\frac{1}{n}}-{e}^{\\frac{1}{\\left(n+1\\right)}}\\right][\/latex] converges or diverges. If it converges, find its sum.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558849\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558849\"]\r\n<div id=\"fs-id1169737160551\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169737160558\">Write out the sequence of partial sums to see which terms cancel.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1169737160537\" data-type=\"solution\">\r\n<p id=\"fs-id1169737160539\">[latex]e - 1[\/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\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\/Catp5TwohC8?controls=0&amp;start=192&amp;end=255&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\/5.2.3_192to255_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.2.3\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169370[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1169737160565\" class=\"project\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox tryit\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Activity: Euler\u2019s Constant<\/h3>\r\n<p id=\"fs-id1169737160572\">We have shown that the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges. Here we investigate the behavior of the partial sums [latex]{S}_{k}[\/latex] as [latex]k\\to \\infty [\/latex]. In particular, we show that they behave like the natural logarithm function by showing that there exists a constant [latex]\\gamma [\/latex] such that<\/p>\r\n\r\n<div id=\"fs-id1169737910599\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}-\\text{ln}k\\to \\gamma \\text{as}k\\to \\infty [\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1169737910664\">This constant [latex]\\gamma [\/latex] is known as <span class=\"no-emphasis\" data-type=\"term\">Euler\u2019s constant<\/span>.<\/p>\r\n\r\n<ol id=\"fs-id1169737910676\" type=\"1\">\r\n \t<li>Let [latex]{T}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}-\\text{ln}k[\/latex]. Evaluate [latex]{T}_{k}[\/latex] for various values of [latex]k[\/latex].<\/li>\r\n \t<li>For [latex]{T}_{k}[\/latex] as defined in part 1. show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] converges by using the following steps.\r\n<ol id=\"fs-id1169738233611\" type=\"a\">\r\n \t<li>Show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] is monotone decreasing. (<em data-effect=\"italics\">Hint:<\/em> Show that [latex]\\text{ln}\\left(1+\\frac{1}{k}&gt;\\frac{1}{\\left(k+1\\right)}\\right)[\/latex]<\/li>\r\n \t<li>Show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] is bounded below by zero. (<em data-effect=\"italics\">Hint:<\/em> Express [latex]\\text{ln}k[\/latex] as a definite integral.)<\/li>\r\n \t<li>Use the Monotone Convergence Theorem to conclude that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] converges. The limit [latex]\\gamma [\/latex] is Euler\u2019s constant.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t<li>Now estimate how far [latex]{T}_{k}[\/latex] is from [latex]\\gamma [\/latex] for a given integer [latex]k[\/latex]. Prove that for [latex]k\\ge 1[\/latex], [latex]0&lt;{T}_{k}-\\gamma \\le \\frac{1}{k}[\/latex] by using the following steps.\r\n<ol id=\"fs-id1169738226116\" type=\"a\">\r\n \t<li>Show that [latex]\\text{ln}\\left(k+1\\right)-\\text{ln}k&lt;\\frac{1}{k}[\/latex].<\/li>\r\n \t<li>Use the result from part a. to show that for any integer [latex]k[\/latex], <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169738226173\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{k+1}&lt;\\frac{1}{k}-\\frac{1}{k+1}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>For any integers [latex]k[\/latex] and [latex]j[\/latex] such that [latex]j&gt;k[\/latex], express [latex]{T}_{k}-{T}_{j}[\/latex] as a telescoping sum by writing<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737214924\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{j}=\\left({T}_{k}-{T}_{k+1}\\right)+\\left({T}_{k+1}-{T}_{k+2}\\right)+\\left({T}_{k+2}-{T}_{k+3}\\right)+\\cdots +\\left({T}_{j - 1}-{T}_{j}\\right)[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nUse the result from part b. combined with this telescoping sum to conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737358876\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{j}&lt;\\frac{1}{k}-\\frac{1}{j}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Apply the limit to both sides of the inequality in part c. to conclude that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1169737358917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-\\gamma \\le \\frac{1}{k}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Estimate [latex]\\gamma [\/latex] to an accuracy of within [latex]0.001[\/latex].<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1169737174590\" 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>Evaluate a telescoping series<\/li>\n<\/ul>\n<\/div>\n<p>Consider the series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n\\left(n+1\\right)}[\/latex]. We discussed this series in the example: Evaluating Limits of Sequences of Partial Sums, showing that the series converges by writing out the first several partial sums [latex]{S}_{1},{S}_{2}\\text{,}\\ldots,{S}_{6}[\/latex] and noticing that they are all of the form [latex]{S}_{k}=\\frac{k}{k+1}[\/latex]. Here we use a different technique to show that this series converges. By using partial fractions, we can write<\/p>\n<div id=\"fs-id1169738154896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{n\\left(n+1\\right)}=\\frac{1}{n}-\\frac{1}{n+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738216460\">Therefore, the series can be written as<\/p>\n<div id=\"fs-id1169738216464\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\frac{1}{n}-\\frac{1}{n+1}\\right]=\\left(1+\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738191325\">Writing out the first several terms in the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex], we see that<\/p>\n<div id=\"fs-id1169738191343\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}=1-\\frac{1}{2}\\hfill \\\\ {S}_{2}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)=1-\\frac{1}{3}\\hfill \\\\ {S}_{3}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)=1-\\frac{1}{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737910267\">In general,<\/p>\n<div id=\"fs-id1169737910270\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\left(1-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\left(\\frac{1}{3}-\\frac{1}{4}\\right)+\\cdots +\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{k+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169738115213\">We notice that the middle terms cancel each other out, leaving only the first and last terms. In a sense, the series collapses like a spyglass with tubes that disappear into each other to shorten the telescope. For this reason, we call a series that has this property a telescoping series. For this series, since [latex]{S}_{k}=1 - \\frac{1}{\\left(k+1\\right)}[\/latex] and [latex]\\frac{1}{\\left(k+1\\right)}\\to 0[\/latex] as [latex]k\\to \\infty[\/latex], the sequence of partial sums converges to [latex]1[\/latex], and therefore the series converges to [latex]1[\/latex].<\/p>\n<div id=\"fs-id1169738077699\" 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-id1169738077703\">A <strong>telescoping series<\/strong> is a series in which most of the terms cancel in each of the partial sums, leaving only some of the first terms and some of the last terms.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169738077712\">For example, any series of the form<\/p>\n<div id=\"fs-id1169738077715\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)+\\cdots[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737394662\">is a telescoping series. We can see this by writing out some of the partial sums. In particular, we see that<\/p>\n<div id=\"fs-id1169737394667\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{l}{S}_{1}={b}_{1}-{b}_{2}\\hfill \\\\ {S}_{2}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)={b}_{1}-{b}_{3}\\hfill \\\\ {S}_{3}=\\left({b}_{1}-{b}_{2}\\right)+\\left({b}_{2}-{b}_{3}\\right)+\\left({b}_{3}-{b}_{4}\\right)={b}_{1}-{b}_{4}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737168045\">In general, the <em data-effect=\"italics\">k<\/em>th partial sum of this series is<\/p>\n<div id=\"fs-id1169737168054\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}={b}_{1}-{b}_{k+1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737168089\">Since the <em data-effect=\"italics\">k<\/em>th partial sum can be simplified to the difference of these two terms, the sequence of partial sums [latex]\\left\\{{S}_{k}\\right\\}[\/latex] will converge if and only if the sequence [latex]\\left\\{{b}_{k+1}\\right\\}[\/latex] converges. Moreover, if the sequence [latex]{b}_{k+1}[\/latex] converges to some finite number [latex]B[\/latex], then the sequence of partial sums converges to [latex]{b}_{1}-B[\/latex], and therefore<\/p>\n<div id=\"fs-id1169738180671\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{b}_{n}-{b}_{n+1}\\right]={b}_{1}-B[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737264635\">In the next example, we show how to use these ideas to analyze a telescoping series of this form.<\/p>\n<div id=\"fs-id1169737264640\" data-type=\"example\">\n<div id=\"fs-id1169737264642\" data-type=\"exercise\">\n<div id=\"fs-id1169737264644\" data-type=\"problem\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Evaluating a Telescoping Series<\/h3>\n<div id=\"fs-id1169737264644\" data-type=\"problem\">\n<p id=\"fs-id1169737264649\">Determine whether the telescoping series<\/p>\n<div id=\"fs-id1169737264652\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right][\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737154216\">converges or diverges. If it converges, find its sum.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Show Solution<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737154220\" data-type=\"solution\">\n<p id=\"fs-id1169737154222\">By writing out terms in the sequence of partial sums, we can see that<\/p>\n<div id=\"fs-id1169737154225\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {S}_{1}& =\\hfill & \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\hfill \\\\ \\hfill {S}_{2}& =\\hfill & \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{3}\\right)\\hfill \\\\ \\hfill {S}_{3}& =\\hfill & \\left(\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{2}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{2}\\right)-\\cos\\left(\\frac{1}{3}\\right)\\right)+\\left(\\cos\\left(\\frac{1}{3}\\right)-\\cos\\left(\\frac{1}{4}\\right)\\right)\\hfill \\\\ & =\\hfill & \\cos\\left(1\\right)-\\cos\\left(\\frac{1}{4}\\right).\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737392321\">In general,<\/p>\n<div id=\"fs-id1169737392324\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{S}_{k}=\\cos\\left(1\\right)-\\cos\\left(\\frac{1}{k+1}\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737392372\">Since [latex]\\frac{1}{\\left(k+1\\right)}\\to 0[\/latex] as [latex]k\\to \\infty[\/latex] and [latex]\\cos{x}[\/latex] is a continuous function, [latex]\\cos\\left(\\frac{1}{\\left(k+1\\right)}\\right)\\to \\cos\\left(0\\right)=1[\/latex]. Therefore, we conclude that [latex]{S}_{k}\\to \\cos\\left(1\\right)-1[\/latex]. The telescoping series converges and the sum is given by<\/p>\n<div id=\"fs-id1169738185032\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[\\cos\\left(\\frac{1}{n}\\right)-\\cos\\left(\\frac{1}{n+1}\\right)\\right]=\\cos\\left(1\\right)-1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1169738185120\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1169738185124\" data-type=\"exercise\">\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1169738185126\" data-type=\"problem\">\n<p id=\"fs-id1169738185128\">Determine whether [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\left[{e}^{\\frac{1}{n}}-{e}^{\\frac{1}{\\left(n+1\\right)}}\\right][\/latex] converges or diverges. If it converges, find its sum.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558849\">Hint<\/span><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737160551\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169737160558\">Write out the sequence of partial sums to see which terms cancel.<\/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=\"q44558859\">Show Solution<\/span><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169737160537\" data-type=\"solution\">\n<p id=\"fs-id1169737160539\">[latex]e - 1[\/latex]<\/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\/Catp5TwohC8?controls=0&amp;start=192&amp;end=255&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\/5.2.3_192to255_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.2.3&#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=\"ohm169370\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169370&theme=oea&iframe_resize_id=ohm169370&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1169737160565\" class=\"project\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox tryit\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Activity: Euler\u2019s Constant<\/h3>\n<p id=\"fs-id1169737160572\">We have shown that the harmonic series [latex]\\displaystyle\\sum _{n=1}^{\\infty }\\frac{1}{n}[\/latex] diverges. Here we investigate the behavior of the partial sums [latex]{S}_{k}[\/latex] as [latex]k\\to \\infty[\/latex]. In particular, we show that they behave like the natural logarithm function by showing that there exists a constant [latex]\\gamma[\/latex] such that<\/p>\n<div id=\"fs-id1169737910599\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}-\\text{ln}k\\to \\gamma \\text{as}k\\to \\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1169737910664\">This constant [latex]\\gamma[\/latex] is known as <span class=\"no-emphasis\" data-type=\"term\">Euler\u2019s constant<\/span>.<\/p>\n<ol id=\"fs-id1169737910676\" type=\"1\">\n<li>Let [latex]{T}_{k}=\\displaystyle\\sum _{n=1}^{k}\\frac{1}{n}-\\text{ln}k[\/latex]. Evaluate [latex]{T}_{k}[\/latex] for various values of [latex]k[\/latex].<\/li>\n<li>For [latex]{T}_{k}[\/latex] as defined in part 1. show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] converges by using the following steps.\n<ol id=\"fs-id1169738233611\" type=\"a\">\n<li>Show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] is monotone decreasing. (<em data-effect=\"italics\">Hint:<\/em> Show that [latex]\\text{ln}\\left(1+\\frac{1}{k}>\\frac{1}{\\left(k+1\\right)}\\right)[\/latex]<\/li>\n<li>Show that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] is bounded below by zero. (<em data-effect=\"italics\">Hint:<\/em> Express [latex]\\text{ln}k[\/latex] as a definite integral.)<\/li>\n<li>Use the Monotone Convergence Theorem to conclude that the sequence [latex]\\left\\{{T}_{k}\\right\\}[\/latex] converges. The limit [latex]\\gamma[\/latex] is Euler\u2019s constant.<\/li>\n<\/ol>\n<\/li>\n<li>Now estimate how far [latex]{T}_{k}[\/latex] is from [latex]\\gamma[\/latex] for a given integer [latex]k[\/latex]. Prove that for [latex]k\\ge 1[\/latex], [latex]0<{T}_{k}-\\gamma \\le \\frac{1}{k}[\/latex] by using the following steps.\n\n\n<ol id=\"fs-id1169738226116\" type=\"a\">\n<li>Show that [latex]\\text{ln}\\left(k+1\\right)-\\text{ln}k<\\frac{1}{k}[\/latex].<\/li>\n<li>Use the result from part a. to show that for any integer [latex]k[\/latex], <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169738226173\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{k+1}<\\frac{1}{k}-\\frac{1}{k+1}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>For any integers [latex]k[\/latex] and [latex]j[\/latex] such that [latex]j>k[\/latex], express [latex]{T}_{k}-{T}_{j}[\/latex] as a telescoping sum by writing<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737214924\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{j}=\\left({T}_{k}-{T}_{k+1}\\right)+\\left({T}_{k+1}-{T}_{k+2}\\right)+\\left({T}_{k+2}-{T}_{k+3}\\right)+\\cdots +\\left({T}_{j - 1}-{T}_{j}\\right)[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nUse the result from part b. combined with this telescoping sum to conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737358876\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-{T}_{j}<\\frac{1}{k}-\\frac{1}{j}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Apply the limit to both sides of the inequality in part c. to conclude that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1169737358917\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}_{k}-\\gamma \\le \\frac{1}{k}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Estimate [latex]\\gamma[\/latex] to an accuracy of within [latex]0.001[\/latex].<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1169737174590\" 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-1700\">\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>5.2.3. <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":10,"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\":\"5.2.3\",\"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-1700","chapter","type-chapter","status-publish","hentry"],"part":160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1700","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\/1700\/revisions"}],"predecessor-version":[{"id":2214,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1700\/revisions\/2214"}],"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\/1700\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1700"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1700"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1700"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}