{"id":864,"date":"2025-06-20T17:18:46","date_gmt":"2025-06-20T17:18:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=864"},"modified":"2025-09-09T19:50:25","modified_gmt":"2025-09-09T19:50:25","slug":"basics-of-differential-equations-fresh-take-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-fresh-take-2\/","title":{"raw":"Sequences and Their Properties: Fresh Take","rendered":"Sequences and Their Properties: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find the pattern in a sequence and write a formula for its terms<\/li>\r\n \t<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\r\n \t<li>Figure out if a sequence converges or diverges<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>What is a Sequence?<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">A sequence is an ordered list of numbers where position matters and patterns help us predict what comes next. Unlike sets where order doesn't matter, sequences are like mathematical \"playlists\"\u2014each number has a specific spot in line.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">We write sequences as [latex]{a_n}[\/latex] where [latex]a_n[\/latex] is the nth term. The sequence [latex]2, 4, 8, 16, 32, \\ldots[\/latex] has [latex]a_1 = 2[\/latex], [latex]a_2 = 4[\/latex], [latex]a_3 = 8[\/latex], and so on.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\"><strong>Two ways to describe sequences:<\/strong><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Explicit formula:<\/strong> Gives you [latex]a_n[\/latex] directly in terms of [latex]n[\/latex]. Want the 50th term? Just plug in [latex]n = 50[\/latex]. For our example: [latex]a_n = 2^n[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Recurrence relation:<\/strong> Defines each term using previous terms. You need a starting value and a rule. For our example: [latex]a_1 = 2[\/latex] and [latex]a_n = 2a_{n-1}[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Since sequences assign exactly one value to each positive integer, they're functions with domain consisting of positive integers. This means we can graph them as isolated points [latex](n, a_n)[\/latex] rather than connected curves.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div>\r\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\r\n<p class=\"whitespace-normal break-words\">Sequences don't have to start at [latex]n = 1[\/latex]\u2014they can begin at [latex]n = 0[\/latex] or any other integer depending on the context.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Special Types of Sequences<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Two fundamental sequence patterns appear frequently and have special names because of their predictable, useful structures.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Arithmetic sequences:<\/strong> Add the same amount each time. The explicit form is always [latex]a_n = cn + b[\/latex] (linear function).<\/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\">Common difference: [latex]d = a_{n+1} - a_n[\/latex] stays constant<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Example: [latex]3, 7, 11, 15, 19, \\ldots[\/latex] has [latex]d = 4[\/latex] and [latex]a_n = 4n - 1[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Geometric sequences:<\/strong> Multiply by the same amount each time. The explicit form is always [latex]a_n = cr^n[\/latex] (exponential function).<\/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\">Common ratio: [latex]r = \\frac{a_{n+1}}{a_n}[\/latex] stays constant<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Example: [latex]2, -\\frac{2}{3}, \\frac{2}{9}, -\\frac{2}{27}, \\ldots[\/latex] has [latex]r = -\\frac{1}{3}[\/latex] and [latex]a_n = 2(-\\frac{1}{3})^{n-1}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Recognition strategy:<\/strong><\/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\">Check differences between consecutive terms (if constant \u2192 arithmetic)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Check ratios between consecutive terms (if constant \u2192 geometric)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">For complex sequences, examine numerators and denominators separately, watch for alternating signs<\/li>\r\n<\/ul>\r\nWrite out several terms, identify the underlying structure, then build either an explicit formula or recurrence relation based on what you observe.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\r\n<p id=\"fs-id1169739242722\">Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence [latex]\\left\\{\\frac{1}{5},-\\frac{1}{7},\\frac{1}{9},-\\frac{1}{11}\\text{,}\\ldots\\right\\}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1169738880222\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739045640\">The denominators form an arithmetic sequence.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1169739093464\" data-type=\"solution\">\r\n<p id=\"fs-id1169738966727\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n+1}}{3+2n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=285&amp;end=348&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.1_285to348_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\r\n<p id=\"fs-id1169739214209\">Find an explicit formula for the sequence defined recursively such that [latex]{a}_{1}=-4[\/latex] and [latex]{a}_{n}={a}_{n - 1}+6[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1169739302833\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739102236\">This is an arithmetic sequence.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1169736854641\" data-type=\"solution\">\r\n<p id=\"fs-id1169739242787\">[latex]{a}_{n}=6n - 10[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=475&amp;end=506&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.1_475to506_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2>Limit of a Sequence<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The most important question about any sequence is: What happens to the terms as [latex]n[\/latex] gets really large? This \"long-term behavior\" determines whether a sequence is useful for mathematical modeling and tells us if the sequence settles down or keeps changing forever.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The fundamental classification:<\/strong><\/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\"><strong>Convergent sequences:<\/strong> Terms get arbitrarily close to some finite number [latex]L[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Divergent sequences:<\/strong> Terms don't settle down to any finite value<\/li>\r\n<\/ul>\r\nConvergence depends only on what happens eventually. You can add, remove, or change any finite number of terms at the beginning without affecting whether the sequence converges.\r\n\r\nSequences can diverge by growing without bound ([latex]\\lim_{n \\to \\infty} a_n = \\infty[\/latex]) or by oscillating forever. When we write [latex]\\lim_{n \\to \\infty} a_n = \\infty[\/latex], we're still saying the sequence diverges\u2014we're just describing how it diverges.\r\n<p class=\"whitespace-normal break-words\"><strong>The formal definition:<\/strong> A sequence [latex]{a_n}[\/latex] converges to [latex]L[\/latex] if for every tiny distance [latex]\\epsilon &gt; 0[\/latex], there exists some point [latex]N[\/latex] such that all terms beyond [latex]N[\/latex] are within [latex]\\epsilon[\/latex] of [latex]L[\/latex].<\/p>\r\nIf [latex]a_n = f(n)[\/latex] for some function [latex]f[\/latex], and you know [latex]\\lim_{x \\to \\infty} f(x) = L[\/latex], then [latex]\\lim_{n \\to \\infty} a_n = L[\/latex]. This lets you use familiar function limit techniques on sequences.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\"><strong>Examples of different behaviors:<\/strong><\/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]{1 - (\\frac{1}{2})^n}[\/latex] converges to 1 (terms: [latex]\\frac{1}{2}, \\frac{3}{4}, \\frac{7}{8}, \\frac{15}{16}, \\ldots[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{1 + 3n}[\/latex] diverges to [latex]\\infty[\/latex] (terms: [latex]4, 7, 10, 13, \\ldots[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{(-1)^n}[\/latex] oscillates between -1 and 1 forever (divergent)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]{\\frac{(-1)^n}{n}}[\/latex] alternates signs but converges to 0<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"text-2xl font-bold mt-1 text-text-100\">Evaluating Sequence Limits<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Once you understand the theory of sequence limits, the real skill lies in choosing the right technique to evaluate them efficiently. Most sequence limits can be found using a combination of basic patterns and algebraic manipulation.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Geometric sequence limits (your foundation):<\/strong><\/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\">If [latex]0 &lt; r &lt; 1[\/latex]: [latex]r^n \\to 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]r = 1[\/latex]: [latex]r^n \\to 1[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]r &gt; 1[\/latex]: [latex]r^n \\to \\infty[\/latex] (diverges)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Key building block:<\/strong> [latex]\\lim_{n \\to \\infty} \\frac{1}{n^k} = 0[\/latex] for any positive integer [latex]k[\/latex]. This simple fact handles many sequences involving polynomial denominators.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Algebraic limit laws:<\/strong> Just like with functions, you can break complex sequences apart:<\/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\">Sum\/difference: [latex]\\lim(a_n \\pm b_n) = \\lim a_n \\pm \\lim b_n[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Product: [latex]\\lim(a_n \\cdot b_n) = (\\lim a_n) \\cdot (\\lim b_n)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Quotient: [latex]\\lim(\\frac{a_n}{b_n}) = \\frac{\\lim a_n}{\\lim b_n}[\/latex] (when denominator limit \u2260 0)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Rational expressions strategy:<\/strong> For sequences like [latex]\\frac{3n^4 - 7n^2 + 5}{6 - 4n^4}[\/latex], factor out the highest power of [latex]n[\/latex] from numerator and denominator. The limit depends on which degree is larger, just like with rational functions.<\/p>\r\nFor sequences involving exponentials or indeterminate forms like [latex](1 + \\frac{4}{n})^n[\/latex], treat them as continuous functions and use logarithmic techniques or L'H\u00f4pital's rule.\r\n\r\n<\/div>\r\n<h2>Advanced Techniques for Sequence Limits<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">When basic limit laws and algebraic manipulation aren't enough, these advanced techniques handle the tricky cases that would otherwise leave you stuck.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>L'H\u00f4pital's Rule for sequences:<\/strong> When you get indeterminate forms like [latex]\\frac{\\infty}{\\infty}[\/latex] or [latex]\\frac{0}{0}[\/latex], treat the sequence as a function and apply L'H\u00f4pital's rule. For [latex]{\\frac{5n^2+1}{e^n}}[\/latex], both numerator and denominator approach infinity, so differentiate top and bottom repeatedly until you get a determinate form.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Continuous functions preserve limits:<\/strong> If [latex]{a_n}[\/latex] converges to [latex]L[\/latex] and [latex]f[\/latex] is continuous at [latex]L[\/latex], then [latex]{f(a_n)}[\/latex] converges to [latex]f(L)[\/latex]. This is incredibly useful for sequences involving square roots, trigonometric functions, exponentials, and logarithms.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Squeeze Theorem:<\/strong> When direct calculation is difficult, \"sandwich\" your sequence between two simpler sequences that converge to the same limit. This is particularly powerful for oscillating sequences.<\/p>\r\n<p class=\"whitespace-normal break-words\">Classic squeeze setup: For [latex]{\\frac{\\cos n}{n^2}}[\/latex], use [latex]-1 \\leq \\cos n \\leq 1[\/latex] to get [latex]-\\frac{1}{n^2} \\leq \\frac{\\cos n}{n^2} \\leq \\frac{1}{n^2}[\/latex]. Since both outer sequences approach 0, the middle one must too.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>When to use each technique:<\/strong><\/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\">L'H\u00f4pital's: Indeterminate forms involving exponentials, polynomials, or logarithms<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Continuous functions: When the sequence formula involves standard functions like [latex]\\sin[\/latex], [latex]\\cos[\/latex], [latex]\\sqrt{x}[\/latex], [latex]e^x[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Squeeze Theorem: Oscillating sequences or when bounds are easier to find than exact values<\/li>\r\n<\/ul>\r\nTry the simplest approach first, then escalate to these advanced techniques when needed.\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169739300567\" data-type=\"problem\">\r\n<p id=\"fs-id1169739300569\">Determine if the sequence [latex]\\left\\{\\sqrt{\\frac{2n+1}{3n+5}}\\right\\}[\/latex] converges. If it converges, find its limit.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558879\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1169739300625\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739300631\">Consider the sequence [latex]\\left\\{\\frac{2n+1}{3n+5}\\right\\}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1169739300606\" data-type=\"solution\">\r\n<p id=\"fs-id1169739300608\">The sequence converges, and its limit is [latex]\\sqrt{\\frac{2}{3}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=479&amp;end=519&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.2_479to519_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.2\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169738920755\" data-type=\"problem\">\r\n<p id=\"fs-id1169738920757\">Find [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{2n-\\sin{n}}{n}[\/latex].<\/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-id1169739252658\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739252665\">Use the fact that [latex]-1\\le \\sin{n}\\le 1[\/latex].<\/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-id1169739252650\" data-type=\"solution\">\r\n<p id=\"fs-id1169739252652\">[latex]2[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=601&amp;end=649&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.2_601to649_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.2\" here (opens in new window)<\/a>.\r\n\r\n<\/section>\r\n<h2>Monotone Convergence Theorem<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The Monotone Convergence Theorem provides a practical way to prove sequences converge even when you can't calculate their exact limits. It combines two key concepts: boundedness and monotonicity (consistent direction of change).<\/p>\r\n<p class=\"whitespace-normal break-words\">If a sequence is both bounded and eventually monotone (either consistently increasing or consistently decreasing), then it must converge. No oscillation, no divergence to infinity\u2014it has to settle down to some finite limit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key definitions:<\/strong><\/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\"><strong>Bounded sequence:<\/strong> Has both upper and lower bounds (all terms stay within some finite range)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Monotone sequence:<\/strong> Eventually increases consistently or decreases consistently<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Eventually monotone:<\/strong> The pattern holds from some point forward (early terms can behave differently)<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>The theorem:<\/strong> If [latex]{a_n}[\/latex] is bounded and eventually monotone, then [latex]{a_n}[\/latex] converges.<\/p>\r\n<p class=\"whitespace-normal break-words\">Why does this work?\u00a0Think of a bounded increasing sequence as climbing a ladder with a ceiling. Since you can't go through the ceiling (bounded above) and you keep climbing (increasing), you must approach some highest reachable point. That's your limit.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Important relationships:<\/strong><\/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\">All convergent sequences are bounded (but bounded sequences aren't necessarily convergent)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Unbounded sequences cannot converge<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Monotonicity eliminates oscillation, which is what makes bounded sequences fail to converge<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">This theorem is especially useful for recursively defined sequences and sequences involving factorials or exponentials where direct limit calculation is difficult.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\r\n<p id=\"fs-id1169739202296\">Consider the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that [latex]{a}_{1}=1[\/latex], [latex]{a}_{n}=\\frac{{a}_{n - 1}}{2}[\/latex]. Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558599\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1169739202370\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1169739202376\">Show the sequence is decreasing and bounded below.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1169739202358\" data-type=\"solution\">\r\n<p id=\"fs-id1169739202360\">[latex]0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V-B05oiNKFk?controls=0&amp;start=713&amp;end=822&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.3_713to822_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"5.1.3\" here (opens in new window)<\/a>.<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find the pattern in a sequence and write a formula for its terms<\/li>\n<li>Determine what value a sequence approaches (if it approaches any value at all)<\/li>\n<li>Figure out if a sequence converges or diverges<\/li>\n<\/ul>\n<\/section>\n<h2>What is a Sequence?<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">A sequence is an ordered list of numbers where position matters and patterns help us predict what comes next. Unlike sets where order doesn&#8217;t matter, sequences are like mathematical &#8220;playlists&#8221;\u2014each number has a specific spot in line.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">We write sequences as [latex]{a_n}[\/latex] where [latex]a_n[\/latex] is the nth term. The sequence [latex]2, 4, 8, 16, 32, \\ldots[\/latex] has [latex]a_1 = 2[\/latex], [latex]a_2 = 4[\/latex], [latex]a_3 = 8[\/latex], and so on.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\"><strong>Two ways to describe sequences:<\/strong><\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Explicit formula:<\/strong> Gives you [latex]a_n[\/latex] directly in terms of [latex]n[\/latex]. Want the 50th term? Just plug in [latex]n = 50[\/latex]. For our example: [latex]a_n = 2^n[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Recurrence relation:<\/strong> Defines each term using previous terms. You need a starting value and a rule. For our example: [latex]a_1 = 2[\/latex] and [latex]a_n = 2a_{n-1}[\/latex].<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Since sequences assign exactly one value to each positive integer, they&#8217;re functions with domain consisting of positive integers. This means we can graph them as isolated points [latex](n, a_n)[\/latex] rather than connected curves.<\/p>\n<\/div>\n<\/div>\n<div>\n<div class=\"grid-cols-1 grid gap-2.5 [&amp;_&gt;_*]:min-w-0\">\n<p class=\"whitespace-normal break-words\">Sequences don&#8217;t have to start at [latex]n = 1[\/latex]\u2014they can begin at [latex]n = 0[\/latex] or any other integer depending on the context.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Special Types of Sequences<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Two fundamental sequence patterns appear frequently and have special names because of their predictable, useful structures.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Arithmetic sequences:<\/strong> Add the same amount each time. The explicit form is always [latex]a_n = cn + b[\/latex] (linear function).<\/p>\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\">\n<li class=\"whitespace-normal break-words\">Common difference: [latex]d = a_{n+1} - a_n[\/latex] stays constant<\/li>\n<li class=\"whitespace-normal break-words\">Example: [latex]3, 7, 11, 15, 19, \\ldots[\/latex] has [latex]d = 4[\/latex] and [latex]a_n = 4n - 1[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Geometric sequences:<\/strong> Multiply by the same amount each time. The explicit form is always [latex]a_n = cr^n[\/latex] (exponential function).<\/p>\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\">\n<li class=\"whitespace-normal break-words\">Common ratio: [latex]r = \\frac{a_{n+1}}{a_n}[\/latex] stays constant<\/li>\n<li class=\"whitespace-normal break-words\">Example: [latex]2, -\\frac{2}{3}, \\frac{2}{9}, -\\frac{2}{27}, \\ldots[\/latex] has [latex]r = -\\frac{1}{3}[\/latex] and [latex]a_n = 2(-\\frac{1}{3})^{n-1}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Recognition strategy:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\">Check differences between consecutive terms (if constant \u2192 arithmetic)<\/li>\n<li class=\"whitespace-normal break-words\">Check ratios between consecutive terms (if constant \u2192 geometric)<\/li>\n<li class=\"whitespace-normal break-words\">For complex sequences, examine numerators and denominators separately, watch for alternating signs<\/li>\n<\/ul>\n<p>Write out several terms, identify the underlying structure, then build either an explicit formula or recurrence relation based on what you observe.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169736593628\" data-type=\"problem\">\n<p id=\"fs-id1169739242722\">Find an explicit formula for the [latex]n\\text{th}[\/latex] term of the sequence [latex]\\left\\{\\frac{1}{5},-\\frac{1}{7},\\frac{1}{9},-\\frac{1}{11}\\text{,}\\ldots\\right\\}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169738880222\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739045640\">The denominators form an arithmetic sequence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739093464\" data-type=\"solution\">\n<p id=\"fs-id1169738966727\">[latex]{a}_{n}=\\frac{{\\left(-1\\right)}^{n+1}}{3+2n}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=285&amp;end=348&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.1_285to348_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738975437\" data-type=\"problem\">\n<p id=\"fs-id1169739214209\">Find an explicit formula for the sequence defined recursively such that [latex]{a}_{1}=-4[\/latex] and [latex]{a}_{n}={a}_{n - 1}+6[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739302833\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739102236\">This is an arithmetic sequence.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169736854641\" data-type=\"solution\">\n<p id=\"fs-id1169739242787\">[latex]{a}_{n}=6n - 10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/oduNoQPFDD4?controls=0&amp;start=475&amp;end=506&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.1_475to506_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2>Limit of a Sequence<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The most important question about any sequence is: What happens to the terms as [latex]n[\/latex] gets really large? This &#8220;long-term behavior&#8221; determines whether a sequence is useful for mathematical modeling and tells us if the sequence settles down or keeps changing forever.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The fundamental classification:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\"><strong>Convergent sequences:<\/strong> Terms get arbitrarily close to some finite number [latex]L[\/latex] as [latex]n \\to \\infty[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Divergent sequences:<\/strong> Terms don&#8217;t settle down to any finite value<\/li>\n<\/ul>\n<p>Convergence depends only on what happens eventually. You can add, remove, or change any finite number of terms at the beginning without affecting whether the sequence converges.<\/p>\n<p>Sequences can diverge by growing without bound ([latex]\\lim_{n \\to \\infty} a_n = \\infty[\/latex]) or by oscillating forever. When we write [latex]\\lim_{n \\to \\infty} a_n = \\infty[\/latex], we&#8217;re still saying the sequence diverges\u2014we&#8217;re just describing how it diverges.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The formal definition:<\/strong> A sequence [latex]{a_n}[\/latex] converges to [latex]L[\/latex] if for every tiny distance [latex]\\epsilon > 0[\/latex], there exists some point [latex]N[\/latex] such that all terms beyond [latex]N[\/latex] are within [latex]\\epsilon[\/latex] of [latex]L[\/latex].<\/p>\n<p>If [latex]a_n = f(n)[\/latex] for some function [latex]f[\/latex], and you know [latex]\\lim_{x \\to \\infty} f(x) = L[\/latex], then [latex]\\lim_{n \\to \\infty} a_n = L[\/latex]. This lets you use familiar function limit techniques on sequences.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\"><strong>Examples of different behaviors:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\">[latex]{1 - (\\frac{1}{2})^n}[\/latex] converges to 1 (terms: [latex]\\frac{1}{2}, \\frac{3}{4}, \\frac{7}{8}, \\frac{15}{16}, \\ldots[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{1 + 3n}[\/latex] diverges to [latex]\\infty[\/latex] (terms: [latex]4, 7, 10, 13, \\ldots[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{(-1)^n}[\/latex] oscillates between -1 and 1 forever (divergent)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]{\\frac{(-1)^n}{n}}[\/latex] alternates signs but converges to 0<\/li>\n<\/ul>\n<\/section>\n<h2 class=\"text-2xl font-bold mt-1 text-text-100\">Evaluating Sequence Limits<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Once you understand the theory of sequence limits, the real skill lies in choosing the right technique to evaluate them efficiently. Most sequence limits can be found using a combination of basic patterns and algebraic manipulation.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Geometric sequence limits (your foundation):<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\">If [latex]0 < r < 1[\/latex]: [latex]r^n \\to 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]r = 1[\/latex]: [latex]r^n \\to 1[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]r > 1[\/latex]: [latex]r^n \\to \\infty[\/latex] (diverges)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Key building block:<\/strong> [latex]\\lim_{n \\to \\infty} \\frac{1}{n^k} = 0[\/latex] for any positive integer [latex]k[\/latex]. This simple fact handles many sequences involving polynomial denominators.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Algebraic limit laws:<\/strong> Just like with functions, you can break complex sequences apart:<\/p>\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\">\n<li class=\"whitespace-normal break-words\">Sum\/difference: [latex]\\lim(a_n \\pm b_n) = \\lim a_n \\pm \\lim b_n[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Product: [latex]\\lim(a_n \\cdot b_n) = (\\lim a_n) \\cdot (\\lim b_n)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Quotient: [latex]\\lim(\\frac{a_n}{b_n}) = \\frac{\\lim a_n}{\\lim b_n}[\/latex] (when denominator limit \u2260 0)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Rational expressions strategy:<\/strong> For sequences like [latex]\\frac{3n^4 - 7n^2 + 5}{6 - 4n^4}[\/latex], factor out the highest power of [latex]n[\/latex] from numerator and denominator. The limit depends on which degree is larger, just like with rational functions.<\/p>\n<p>For sequences involving exponentials or indeterminate forms like [latex](1 + \\frac{4}{n})^n[\/latex], treat them as continuous functions and use logarithmic techniques or L&#8217;H\u00f4pital&#8217;s rule.<\/p>\n<\/div>\n<h2>Advanced Techniques for Sequence Limits<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When basic limit laws and algebraic manipulation aren&#8217;t enough, these advanced techniques handle the tricky cases that would otherwise leave you stuck.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>L&#8217;H\u00f4pital&#8217;s Rule for sequences:<\/strong> When you get indeterminate forms like [latex]\\frac{\\infty}{\\infty}[\/latex] or [latex]\\frac{0}{0}[\/latex], treat the sequence as a function and apply L&#8217;H\u00f4pital&#8217;s rule. For [latex]{\\frac{5n^2+1}{e^n}}[\/latex], both numerator and denominator approach infinity, so differentiate top and bottom repeatedly until you get a determinate form.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Continuous functions preserve limits:<\/strong> If [latex]{a_n}[\/latex] converges to [latex]L[\/latex] and [latex]f[\/latex] is continuous at [latex]L[\/latex], then [latex]{f(a_n)}[\/latex] converges to [latex]f(L)[\/latex]. This is incredibly useful for sequences involving square roots, trigonometric functions, exponentials, and logarithms.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Squeeze Theorem:<\/strong> When direct calculation is difficult, &#8220;sandwich&#8221; your sequence between two simpler sequences that converge to the same limit. This is particularly powerful for oscillating sequences.<\/p>\n<p class=\"whitespace-normal break-words\">Classic squeeze setup: For [latex]{\\frac{\\cos n}{n^2}}[\/latex], use [latex]-1 \\leq \\cos n \\leq 1[\/latex] to get [latex]-\\frac{1}{n^2} \\leq \\frac{\\cos n}{n^2} \\leq \\frac{1}{n^2}[\/latex]. Since both outer sequences approach 0, the middle one must too.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>When to use each technique:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\">L&#8217;H\u00f4pital&#8217;s: Indeterminate forms involving exponentials, polynomials, or logarithms<\/li>\n<li class=\"whitespace-normal break-words\">Continuous functions: When the sequence formula involves standard functions like [latex]\\sin[\/latex], [latex]\\cos[\/latex], [latex]\\sqrt{x}[\/latex], [latex]e^x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Squeeze Theorem: Oscillating sequences or when bounds are easier to find than exact values<\/li>\n<\/ul>\n<p>Try the simplest approach first, then escalate to these advanced techniques when needed.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169739300567\" data-type=\"problem\">\n<p id=\"fs-id1169739300569\">Determine if the sequence [latex]\\left\\{\\sqrt{\\frac{2n+1}{3n+5}}\\right\\}[\/latex] converges. If it converges, find its limit.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558879\">Hint<\/button><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739300625\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739300631\">Consider the sequence [latex]\\left\\{\\frac{2n+1}{3n+5}\\right\\}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558889\">Show Solution<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739300606\" data-type=\"solution\">\n<p id=\"fs-id1169739300608\">The sequence converges, and its limit is [latex]\\sqrt{\\frac{2}{3}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=479&amp;end=519&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.2_479to519_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.2&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169738920755\" data-type=\"problem\">\n<p id=\"fs-id1169738920757\">Find [latex]\\underset{n\\to \\infty }{\\text{lim}}\\frac{2n-\\sin{n}}{n}[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558849\">Hint<\/button><\/p>\n<div id=\"q44558849\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739252658\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739252665\">Use the fact that [latex]-1\\le \\sin{n}\\le 1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739252650\" data-type=\"solution\">\n<p id=\"fs-id1169739252652\">[latex]2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/6Wri3AvC6xg?controls=0&amp;start=601&amp;end=649&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/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\/5.1.2_601to649_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.2&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<h2>Monotone Convergence Theorem<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The Monotone Convergence Theorem provides a practical way to prove sequences converge even when you can&#8217;t calculate their exact limits. It combines two key concepts: boundedness and monotonicity (consistent direction of change).<\/p>\n<p class=\"whitespace-normal break-words\">If a sequence is both bounded and eventually monotone (either consistently increasing or consistently decreasing), then it must converge. No oscillation, no divergence to infinity\u2014it has to settle down to some finite limit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key definitions:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\"><strong>Bounded sequence:<\/strong> Has both upper and lower bounds (all terms stay within some finite range)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Monotone sequence:<\/strong> Eventually increases consistently or decreases consistently<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Eventually monotone:<\/strong> The pattern holds from some point forward (early terms can behave differently)<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>The theorem:<\/strong> If [latex]{a_n}[\/latex] is bounded and eventually monotone, then [latex]{a_n}[\/latex] converges.<\/p>\n<p class=\"whitespace-normal break-words\">Why does this work?\u00a0Think of a bounded increasing sequence as climbing a ladder with a ceiling. Since you can&#8217;t go through the ceiling (bounded above) and you keep climbing (increasing), you must approach some highest reachable point. That&#8217;s your limit.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Important relationships:<\/strong><\/p>\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\">\n<li class=\"whitespace-normal break-words\">All convergent sequences are bounded (but bounded sequences aren&#8217;t necessarily convergent)<\/li>\n<li class=\"whitespace-normal break-words\">Unbounded sequences cannot converge<\/li>\n<li class=\"whitespace-normal break-words\">Monotonicity eliminates oscillation, which is what makes bounded sequences fail to converge<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">This theorem is especially useful for recursively defined sequences and sequences involving factorials or exponentials where direct limit calculation is difficult.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1169739202294\" data-type=\"problem\">\n<p id=\"fs-id1169739202296\">Consider the sequence [latex]\\left\\{{a}_{n}\\right\\}[\/latex] defined recursively such that [latex]{a}_{1}=1[\/latex], [latex]{a}_{n}=\\frac{{a}_{n - 1}}{2}[\/latex]. Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558599\">Hint<\/button><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739202370\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1169739202376\">Show the sequence is decreasing and bounded below.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558699\">Show Solution<\/button><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1169739202358\" data-type=\"solution\">\n<p id=\"fs-id1169739202360\">[latex]0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/V-B05oiNKFk?controls=0&amp;start=713&amp;end=822&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/5.1.3_713to822_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;5.1.3&#8221; here (opens in new window)<\/a>.<\/section>\n","protected":false},"author":15,"menu_order":11,"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\/864"}],"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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/864\/revisions"}],"predecessor-version":[{"id":2261,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/864\/revisions\/2261"}],"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\/864\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=864"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=864"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=864"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}