{"id":875,"date":"2025-06-20T17:19:22","date_gmt":"2025-06-20T17:19:22","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=875"},"modified":"2025-09-18T14:47:37","modified_gmt":"2025-09-18T14:47:37","slug":"direction-fields-and-eulers-method-apply-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/direction-fields-and-eulers-method-apply-it-2\/","title":{"raw":"Introduction to Series: Apply It","rendered":"Introduction to Series: Apply It"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Understand what we mean by the sum of an infinite series<\/li>\r\n \t<li>Find the sum of a geometric series<\/li>\r\n \t<li>Calculate the sum of a telescoping series<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Euler's Constant: Connecting the Harmonic Series to the Natural Logarithm<\/h2>\r\nThe harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] diverges, but it does so in a fascinating way. While its partial sums grow without bound, they grow very slowly\u2014approximately like the natural logarithm function. This connection leads us to one of mathematics' most important constants.\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 Euler's constant, and its discovery reveals deep connections between discrete sums and continuous integrals\u2014a bridge between different areas of mathematics that continues to inspire research today.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"129609\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"129609\"]\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 1[\/latex]: [latex]T_1 = \\frac{1}{1} - \\ln(1) = 1 - 0 = 1[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 2[\/latex]: [latex]T_2 = \\left(\\frac{1}{1} + \\frac{1}{2}\\right) - \\ln(2) = 1.5 - 0.693147... \\approx 0.806853[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 3[\/latex]: [latex]T_3 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3}\\right) - \\ln(3) = 1.833333... - 1.098612... \\approx 0.734721[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 4[\/latex]: [latex]T_4 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4}\\right) - \\ln(4) = 2.083333... - 1.386294... \\approx 0.697039[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 5[\/latex]: [latex]T_5 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5}\\right) - \\ln(5) = 2.283333... - 1.609438... \\approx 0.673895[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 10[\/latex]: [latex]T_{10} = \\sum_{n=1}^{10}\\frac{1}{n} - \\ln(10) = 2.928968... - 2.302585... \\approx 0.626383[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 100[\/latex]: [latex]T_{100} = \\sum_{n=1}^{100}\\frac{1}{n} - \\ln(100) = 5.187378... - 4.605170... \\approx 0.582207[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Notice that [latex]T_k[\/latex] appears to be decreasing and approaching a limit around [latex]0.577.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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[reveal-answer q=\"487281\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"487281\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">To show [latex]{T_k}[\/latex] is monotone decreasing, we need to show [latex]T_{k+1} &lt; T_k[\/latex], or equivalently, [latex]T_k - T_{k+1} &gt; 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]T_k - T_{k+1} = \\left(\\sum_{n=1}^{k}\\frac{1}{n} - \\ln k\\right) - \\left(\\sum_{n=1}^{k+1}\\frac{1}{n} - \\ln(k+1)\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]= \\sum_{n=1}^{k}\\frac{1}{n} - \\ln k - \\sum_{n=1}^{k+1}\\frac{1}{n} + \\ln(k+1)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]= -\\frac{1}{k+1} + \\ln(k+1) - \\ln k[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]= \\ln\\left(\\frac{k+1}{k}\\right) - \\frac{1}{k+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">[latex]= \\ln\\left(1 + \\frac{1}{k}\\right) - \\frac{1}{k+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Now we need to show that [latex]\\ln\\left(1 + \\frac{1}{k}\\right) &gt; \\frac{1}{k+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Let [latex]f(x) = \\ln(1+x) - \\frac{x}{1+x}[\/latex] for [latex]x &gt; 0[\/latex]. We need to show [latex]f\\left(\\frac{1}{k}\\right) &gt; 0[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Taking the derivative: [latex]f'(x) = \\frac{1}{1+x} - \\frac{(1+x) - x}{(1+x)^2} = \\frac{1}{1+x} - \\frac{1}{(1+x)^2} = \\frac{1+x-1}{(1+x)^2} = \\frac{x}{(1+x)^2}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]x &gt; 0[\/latex], we have [latex]f'(x) &gt; 0[\/latex], so [latex]f(x)[\/latex] is increasing for [latex]x &gt; 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Also, [latex]f(0) = \\ln(1) - 0 = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, for [latex]x &gt; 0[\/latex], we have [latex]f(x) &gt; f(0) = 0[\/latex], which means [latex]\\ln(1+x) &gt; \\frac{x}{1+x}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Setting [latex]x = \\frac{1}{k}[\/latex], we get [latex]\\ln\\left(1 + \\frac{1}{k}\\right) &gt; \\frac{1}{k+1}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]T_k - T_{k+1} &gt; 0[\/latex], so [latex]{T_k}[\/latex] is monotone decreasing.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">We can express [latex]\\ln k = \\int_1^k \\frac{1}{x} dx[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Now, consider the Riemann sum approximation using right endpoints: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n+1} \\cdot 1 = \\sum_{n=2}^{k} \\frac{1}{n}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is a right Riemann sum for [latex]\\int_1^k \\frac{1}{x} dx[\/latex] with subintervals [latex][n, n+1][\/latex] for [latex]n = 1, 2, ..., k-1[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x) = \\frac{1}{x}[\/latex] is decreasing on [latex][1,k][\/latex], the right Riemann sum underestimates the integral: [latex]\\sum_{n=2}^{k} \\frac{1}{n} &lt; \\int_1^k \\frac{1}{x} dx = \\ln k[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\sum_{n=1}^{k} \\frac{1}{n} = 1 + \\sum_{n=2}^{k} \\frac{1}{n} &lt; 1 + \\ln k[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">This gives us: [latex]T_k = \\sum_{n=1}^{k} \\frac{1}{n} - \\ln k &lt; 1 + \\ln k - \\ln k = 1[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For a tighter bound, using left Riemann sums: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n} \\cdot 1 = \\sum_{n=1}^{k-1} \\frac{1}{n}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x) = \\frac{1}{x}[\/latex] is decreasing, the left Riemann sum overestimates the integral: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n} &gt; \\int_1^k \\frac{1}{x} dx = \\ln k[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\sum_{n=1}^{k} \\frac{1}{n} = \\sum_{n=1}^{k-1} \\frac{1}{n} + \\frac{1}{k} &gt; \\ln k + \\frac{1}{k}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">This gives us: [latex]T_k = \\sum_{n=1}^{k} \\frac{1}{n} - \\ln k &gt; \\ln k + \\frac{1}{k} - \\ln k = \\frac{1}{k} &gt; 0[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]T_k &gt; 0[\/latex] for all [latex]k \\geq 1[\/latex], so [latex]{T_k}[\/latex] is bounded below by zero.<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">From parts (a) and (b), we have shown that:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">The sequence [latex]{T_k}[\/latex] is monotone decreasing<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The sequence [latex]{T_k}[\/latex] is bounded below by zero<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">By the Monotone Convergence Theorem, since [latex]{T_k}[\/latex] is monotone decreasing and bounded below, the sequence [latex]{T_k}[\/latex] converges to some limit.<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">We define this limit as Euler's constant: [latex]\\gamma = \\lim_{k \\to \\infty} T_k = \\lim_{k \\to \\infty} \\left(\\sum_{n=1}^{k}\\frac{1}{n} - \\ln k\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\gamma \\approx 0.5772156649...[\/latex] (Euler's constant).<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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><\/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\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><\/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><\/li>\r\n \t<li>Estimate [latex]\\gamma [\/latex] to an accuracy of within [latex]0.001[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"217991\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"217991\"]\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">We need to show [latex]\\ln(k+1) - \\ln k &lt; \\frac{1}{k}[\/latex], which is equivalent to [latex]\\ln\\left(\\frac{k+1}{k}\\right) &lt; \\frac{1}{k}[\/latex], or [latex]\\ln\\left(1 + \\frac{1}{k}\\right) &lt; \\frac{1}{k}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Consider the function [latex]g(x) = \\ln(1+x) - x[\/latex] for [latex]x &gt; 0[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Taking the derivative: [latex]g'(x) = \\frac{1}{1+x} - 1 = \\frac{1-(1+x)}{1+x} = \\frac{-x}{1+x}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Since [latex]x &gt; 0[\/latex], we have [latex]g'(x) &lt; 0[\/latex], so [latex]g(x)[\/latex] is decreasing for [latex]x &gt; 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Also, [latex]g(0) = \\ln(1) - 0 = 0[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, for [latex]x &gt; 0[\/latex], we have [latex]g(x) &lt; g(0) = 0[\/latex], which means [latex]\\ln(1+x) &lt; x[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Setting [latex]x = \\frac{1}{k}[\/latex], we get [latex]\\ln\\left(1 + \\frac{1}{k}\\right) &lt; \\frac{1}{k}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\ln(k+1) - \\ln k &lt; \\frac{1}{k}[\/latex].<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-pre-wrap break-words\">From the solution to Problem 2(a), we found: [latex]T_k - T_{k+1} = \\ln(k+1) - \\ln k - \\frac{1}{k+1}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">From part (a), we know [latex]\\ln(k+1) - \\ln k &lt; \\frac{1}{k}[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]T_k - T_{k+1} = \\ln(k+1) - \\ln k - \\frac{1}{k+1} &lt; \\frac{1}{k} - \\frac{1}{k+1}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-pre-wrap break-words\">We can write: [latex]T_k - T_j = (T_k - T_{k+1}) + (T_{k+1} - T_{k+2}) + (T_{k+2} - T_{k+3}) + \\cdots + (T_{j-1} - T_j)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This is a telescoping sum. From part (b), each term satisfies:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]T_k - T_{k+1} &lt; \\frac{1}{k} - \\frac{1}{k+1}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]T_{k+1} - T_{k+2} &lt; \\frac{1}{k+1} - \\frac{1}{k+2}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]T_{k+2} - T_{k+3} &lt; \\frac{1}{k+2} - \\frac{1}{k+3}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\dots[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]T_{j-1} - T_j &lt; \\frac{1}{j-1} - \\frac{1}{j}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">Adding all these inequalities: [latex]T_k - T_j &lt; \\left(\\frac{1}{k} - \\frac{1}{k+1}\\right) + \\left(\\frac{1}{k+1} - \\frac{1}{k+2}\\right) + \\cdots + \\left(\\frac{1}{j-1} - \\frac{1}{j}\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The right side is a telescoping sum that simplifies to: [latex]T_k - T_j &lt; \\frac{1}{k} - \\frac{1}{j}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">From part (c), we have [latex]T_k - T_j &lt; \\frac{1}{k} - \\frac{1}{j}[\/latex] for any [latex]j &gt; k[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Taking the limit as [latex]j \\to \\infty[\/latex]: [latex]\\lim_{j \\to \\infty} (T_k - T_j) \\leq \\lim_{j \\to \\infty} \\left(\\frac{1}{k} - \\frac{1}{j}\\right)[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]\\lim_{j \\to \\infty} T_j = \\gamma[\/latex] and [latex]\\lim_{j \\to \\infty} \\frac{1}{j} = 0[\/latex]: [latex]T_k - \\gamma \\leq \\frac{1}{k} - 0 = \\frac{1}{k}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">We already showed in Problem 2(b) that [latex]T_k &gt; 0[\/latex], and since [latex]{T_k}[\/latex] is decreasing and converges to [latex]\\gamma[\/latex], we have [latex]T_k &gt; \\gamma[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore: [latex]0 &lt; T_k - \\gamma \\leq \\frac{1}{k}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p class=\"whitespace-normal break-words\">From part (d), we know that [latex]|T_k - \\gamma| = T_k - \\gamma \\leq \\frac{1}{k}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">To achieve accuracy within [latex]0.001[\/latex], we need [latex]\\frac{1}{k} \\leq 0.001[\/latex], which means [latex]k \\geq 1000[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Let's calculate [latex]T_{1000}[\/latex]: [latex]T_{1000} = \\sum_{n=1}^{1000}\\frac{1}{n} - \\ln(1000)[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Using numerical computation: [latex]\\sum_{n=1}^{1000}\\frac{1}{n} \\approx 7.485470861[\/latex] [latex]\\ln(1000) \\approx 6.907755279[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]T_{1000} \\approx 7.485470861 - 6.907755279 \\approx 0.577715582[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]T_{1000} - \\gamma \\leq \\frac{1}{1000} = 0.001[\/latex], we have: [latex]\\gamma \\geq T_{1000} - 0.001 \\approx 0.577715582 - 0.001 = 0.576715582[\/latex] [latex]\\gamma \\leq T_{1000} \\approx 0.577715582[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Therefore, to an accuracy of within [latex]0.001[\/latex]: [latex]\\gamma \\approx 0.5777[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">The actual value of Euler's constant is [latex]\\gamma \\approx 0.5772156649...[\/latex], which confirms our estimate is accurate to within the required tolerance.<\/p>\r\n<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Understand what we mean by the sum of an infinite series<\/li>\n<li>Find the sum of a geometric series<\/li>\n<li>Calculate the sum of a telescoping series<\/li>\n<\/ul>\n<\/section>\n<h2>Euler&#8217;s Constant: Connecting the Harmonic Series to the Natural Logarithm<\/h2>\n<p>The harmonic series [latex]\\displaystyle\\sum_{n=1}^{\\infty}\\frac{1}{n}[\/latex] diverges, but it does so in a fascinating way. While its partial sums grow without bound, they grow very slowly\u2014approximately like the natural logarithm function. This connection leads us to one of mathematics&#8217; most important constants.<\/p>\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 Euler&#8217;s constant, and its discovery reveals deep connections between discrete sums and continuous integrals\u2014a bridge between different areas of mathematics that continues to inspire research today.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">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].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q129609\">Show Answer<\/button><\/p>\n<div id=\"q129609\" class=\"hidden-answer\" style=\"display: none\">\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 1[\/latex]: [latex]T_1 = \\frac{1}{1} - \\ln(1) = 1 - 0 = 1[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 2[\/latex]: [latex]T_2 = \\left(\\frac{1}{1} + \\frac{1}{2}\\right) - \\ln(2) = 1.5 - 0.693147... \\approx 0.806853[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 3[\/latex]: [latex]T_3 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3}\\right) - \\ln(3) = 1.833333... - 1.098612... \\approx 0.734721[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 4[\/latex]: [latex]T_4 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4}\\right) - \\ln(4) = 2.083333... - 1.386294... \\approx 0.697039[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 5[\/latex]: [latex]T_5 = \\left(\\frac{1}{1} + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5}\\right) - \\ln(5) = 2.283333... - 1.609438... \\approx 0.673895[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 10[\/latex]: [latex]T_{10} = \\sum_{n=1}^{10}\\frac{1}{n} - \\ln(10) = 2.928968... - 2.302585... \\approx 0.626383[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For [latex]k = 100[\/latex]: [latex]T_{100} = \\sum_{n=1}^{100}\\frac{1}{n} - \\ln(100) = 5.187378... - 4.605170... \\approx 0.582207[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Notice that [latex]T_k[\/latex] appears to be decreasing and approaching a limit around [latex]0.577.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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.<\/p>\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<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q487281\">Show Answer<\/button><\/p>\n<div id=\"q487281\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">To show [latex]{T_k}[\/latex] is monotone decreasing, we need to show [latex]T_{k+1} < T_k[\/latex], or equivalently, [latex]T_k - T_{k+1} > 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">[latex]T_k - T_{k+1} = \\left(\\sum_{n=1}^{k}\\frac{1}{n} - \\ln k\\right) - \\left(\\sum_{n=1}^{k+1}\\frac{1}{n} - \\ln(k+1)\\right)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]= \\sum_{n=1}^{k}\\frac{1}{n} - \\ln k - \\sum_{n=1}^{k+1}\\frac{1}{n} + \\ln(k+1)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]= -\\frac{1}{k+1} + \\ln(k+1) - \\ln k[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]= \\ln\\left(\\frac{k+1}{k}\\right) - \\frac{1}{k+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">[latex]= \\ln\\left(1 + \\frac{1}{k}\\right) - \\frac{1}{k+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Now we need to show that [latex]\\ln\\left(1 + \\frac{1}{k}\\right) > \\frac{1}{k+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Let [latex]f(x) = \\ln(1+x) - \\frac{x}{1+x}[\/latex] for [latex]x > 0[\/latex]. We need to show [latex]f\\left(\\frac{1}{k}\\right) > 0[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Taking the derivative: [latex]f'(x) = \\frac{1}{1+x} - \\frac{(1+x) - x}{(1+x)^2} = \\frac{1}{1+x} - \\frac{1}{(1+x)^2} = \\frac{1+x-1}{(1+x)^2} = \\frac{x}{(1+x)^2}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]x > 0[\/latex], we have [latex]f'(x) > 0[\/latex], so [latex]f(x)[\/latex] is increasing for [latex]x > 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Also, [latex]f(0) = \\ln(1) - 0 = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, for [latex]x > 0[\/latex], we have [latex]f(x) > f(0) = 0[\/latex], which means [latex]\\ln(1+x) > \\frac{x}{1+x}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Setting [latex]x = \\frac{1}{k}[\/latex], we get [latex]\\ln\\left(1 + \\frac{1}{k}\\right) > \\frac{1}{k+1}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]T_k - T_{k+1} > 0[\/latex], so [latex]{T_k}[\/latex] is monotone decreasing.<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">We can express [latex]\\ln k = \\int_1^k \\frac{1}{x} dx[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Now, consider the Riemann sum approximation using right endpoints: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n+1} \\cdot 1 = \\sum_{n=2}^{k} \\frac{1}{n}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is a right Riemann sum for [latex]\\int_1^k \\frac{1}{x} dx[\/latex] with subintervals [latex][n, n+1][\/latex] for [latex]n = 1, 2, ..., k-1[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x) = \\frac{1}{x}[\/latex] is decreasing on [latex][1,k][\/latex], the right Riemann sum underestimates the integral: [latex]\\sum_{n=2}^{k} \\frac{1}{n} < \\int_1^k \\frac{1}{x} dx = \\ln k[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\sum_{n=1}^{k} \\frac{1}{n} = 1 + \\sum_{n=2}^{k} \\frac{1}{n} < 1 + \\ln k[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">This gives us: [latex]T_k = \\sum_{n=1}^{k} \\frac{1}{n} - \\ln k < 1 + \\ln k - \\ln k = 1[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For a tighter bound, using left Riemann sums: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n} \\cdot 1 = \\sum_{n=1}^{k-1} \\frac{1}{n}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x) = \\frac{1}{x}[\/latex] is decreasing, the left Riemann sum overestimates the integral: [latex]\\sum_{n=1}^{k-1} \\frac{1}{n} > \\int_1^k \\frac{1}{x} dx = \\ln k[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]\\sum_{n=1}^{k} \\frac{1}{n} = \\sum_{n=1}^{k-1} \\frac{1}{n} + \\frac{1}{k} > \\ln k + \\frac{1}{k}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">This gives us: [latex]T_k = \\sum_{n=1}^{k} \\frac{1}{n} - \\ln k > \\ln k + \\frac{1}{k} - \\ln k = \\frac{1}{k} > 0[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]T_k > 0[\/latex] for all [latex]k \\geq 1[\/latex], so [latex]{T_k}[\/latex] is bounded below by zero.<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">From parts (a) and (b), we have shown that:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">The sequence [latex]{T_k}[\/latex] is monotone decreasing<\/li>\n<li class=\"whitespace-normal break-words\">The sequence [latex]{T_k}[\/latex] is bounded below by zero<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">By the Monotone Convergence Theorem, since [latex]{T_k}[\/latex] is monotone decreasing and bounded below, the sequence [latex]{T_k}[\/latex] converges to some limit.<\/p>\n<p class=\"whitespace-pre-wrap break-words\">We define this limit as Euler&#8217;s constant: [latex]\\gamma = \\lim_{k \\to \\infty} T_k = \\lim_{k \\to \\infty} \\left(\\sum_{n=1}^{k}\\frac{1}{n} - \\ln k\\right)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\gamma \\approx 0.5772156649...[\/latex] (Euler&#8217;s constant).<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<\/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>Use 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<\/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<\/li>\n<li>Estimate [latex]\\gamma[\/latex] to an accuracy of within [latex]0.001[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q217991\">Show Answer<\/button><\/p>\n<div id=\"q217991\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha;\">\n<li>\n<p class=\"whitespace-normal break-words\">We need to show [latex]\\ln(k+1) - \\ln k < \\frac{1}{k}[\/latex], which is equivalent to [latex]\\ln\\left(\\frac{k+1}{k}\\right) < \\frac{1}{k}[\/latex], or [latex]\\ln\\left(1 + \\frac{1}{k}\\right) < \\frac{1}{k}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Consider the function [latex]g(x) = \\ln(1+x) - x[\/latex] for [latex]x > 0[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Taking the derivative: [latex]g'(x) = \\frac{1}{1+x} - 1 = \\frac{1-(1+x)}{1+x} = \\frac{-x}{1+x}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Since [latex]x > 0[\/latex], we have [latex]g'(x) < 0[\/latex], so [latex]g(x)[\/latex] is decreasing for [latex]x > 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Also, [latex]g(0) = \\ln(1) - 0 = 0[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, for [latex]x > 0[\/latex], we have [latex]g(x) < g(0) = 0[\/latex], which means [latex]\\ln(1+x) < x[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Setting [latex]x = \\frac{1}{k}[\/latex], we get [latex]\\ln\\left(1 + \\frac{1}{k}\\right) < \\frac{1}{k}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore, [latex]\\ln(k+1) - \\ln k < \\frac{1}{k}[\/latex].<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-pre-wrap break-words\">From the solution to Problem 2(a), we found: [latex]T_k - T_{k+1} = \\ln(k+1) - \\ln k - \\frac{1}{k+1}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">From part (a), we know [latex]\\ln(k+1) - \\ln k < \\frac{1}{k}[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]T_k - T_{k+1} = \\ln(k+1) - \\ln k - \\frac{1}{k+1} < \\frac{1}{k} - \\frac{1}{k+1}[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-pre-wrap break-words\">We can write: [latex]T_k - T_j = (T_k - T_{k+1}) + (T_{k+1} - T_{k+2}) + (T_{k+2} - T_{k+3}) + \\cdots + (T_{j-1} - T_j)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This is a telescoping sum. From part (b), each term satisfies:<\/p>\n<ul class=\"&#091;&amp;:not(:last-child)_ul&#093;:pb-1 &#091;&amp;:not(:last-child)_ol&#093;:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]T_k - T_{k+1} < \\frac{1}{k} - \\frac{1}{k+1}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_{k+1} - T_{k+2} < \\frac{1}{k+1} - \\frac{1}{k+2}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_{k+2} - T_{k+3} < \\frac{1}{k+2} - \\frac{1}{k+3}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\dots[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]T_{j-1} - T_j < \\frac{1}{j-1} - \\frac{1}{j}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">Adding all these inequalities: [latex]T_k - T_j < \\left(\\frac{1}{k} - \\frac{1}{k+1}\\right) + \\left(\\frac{1}{k+1} - \\frac{1}{k+2}\\right) + \\cdots + \\left(\\frac{1}{j-1} - \\frac{1}{j}\\right)[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">The right side is a telescoping sum that simplifies to: [latex]T_k - T_j < \\frac{1}{k} - \\frac{1}{j}[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">From part (c), we have [latex]T_k - T_j < \\frac{1}{k} - \\frac{1}{j}[\/latex] for any [latex]j > k[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Taking the limit as [latex]j \\to \\infty[\/latex]: [latex]\\lim_{j \\to \\infty} (T_k - T_j) \\leq \\lim_{j \\to \\infty} \\left(\\frac{1}{k} - \\frac{1}{j}\\right)[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]\\lim_{j \\to \\infty} T_j = \\gamma[\/latex] and [latex]\\lim_{j \\to \\infty} \\frac{1}{j} = 0[\/latex]: [latex]T_k - \\gamma \\leq \\frac{1}{k} - 0 = \\frac{1}{k}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">We already showed in Problem 2(b) that [latex]T_k > 0[\/latex], and since [latex]{T_k}[\/latex] is decreasing and converges to [latex]\\gamma[\/latex], we have [latex]T_k > \\gamma[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Therefore: [latex]0 < T_k - \\gamma \\leq \\frac{1}{k}[\/latex]<\/p>\n<\/li>\n<li>\n<p class=\"whitespace-normal break-words\">From part (d), we know that [latex]|T_k - \\gamma| = T_k - \\gamma \\leq \\frac{1}{k}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To achieve accuracy within [latex]0.001[\/latex], we need [latex]\\frac{1}{k} \\leq 0.001[\/latex], which means [latex]k \\geq 1000[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Let&#8217;s calculate [latex]T_{1000}[\/latex]: [latex]T_{1000} = \\sum_{n=1}^{1000}\\frac{1}{n} - \\ln(1000)[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Using numerical computation: [latex]\\sum_{n=1}^{1000}\\frac{1}{n} \\approx 7.485470861[\/latex] [latex]\\ln(1000) \\approx 6.907755279[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore: [latex]T_{1000} \\approx 7.485470861 - 6.907755279 \\approx 0.577715582[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]T_{1000} - \\gamma \\leq \\frac{1}{1000} = 0.001[\/latex], we have: [latex]\\gamma \\geq T_{1000} - 0.001 \\approx 0.577715582 - 0.001 = 0.576715582[\/latex] [latex]\\gamma \\leq T_{1000} \\approx 0.577715582[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Therefore, to an accuracy of within [latex]0.001[\/latex]: [latex]\\gamma \\approx 0.5777[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">The actual value of Euler&#8217;s constant is [latex]\\gamma \\approx 0.5772156649...[\/latex], which confirms our estimate is accurate to within the required tolerance.<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":671,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/875"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/875\/revisions"}],"predecessor-version":[{"id":2377,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/875\/revisions\/2377"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/671"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/875\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=875"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=875"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=875"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=875"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}