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Calculus II: Series Convergence Cheat Sheet
Calculus II: Series Convergence Cheat Sheet
Series Convergence Cheat Sheet
Your guide to conquering infinite series!
Convergence Tests Summary
Test
Conditions & Limit
Conclusion
n-th Term Test for Divergence
\(\lim_{n \to \infty} a_n \neq 0\)
Series Diverges. (Inconclusive if limit is 0).
p-Series\(\sum \frac{1}{n^p}\)
Check the value of \(p\).
Converges if \(p > 1\).Diverges if \(p \le 1\).
Alternating Series Test\(\sum (-1)^n b_n\)
1. \(b_{n+1} \le b_n\) (decreasing)2. \(\lim_{n \to \infty} b_n = 0\)
Series Converges. (Check for absolute convergence separately).
Ratio Test
\(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)
Abs. Convergent if \(L < 1\).Divergent if \(L > 1\).Inconclusive if \(L = 1\).
Root Test
\(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\)
Abs. Convergent if \(L < 1\).Divergent if \(L > 1\).Inconclusive if \(L = 1\).
The Ratio Test
This is a powerful test, especially for series involving factorials ($n!$) or exponentials ($k^n$).
For a series \(\sum a_n\), calculate the limit:
$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
If \(L < 1\), the series is absolutely convergent (and therefore convergent).
If \(L > 1\) or \(L = \infty\), the series is divergent.
If \(L = 1\), the Ratio Test is inconclusive.…
Calculus II: Series Convergence Cheat Sheet
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<header class="text-center mb-12">
<h1 class="text-5xl md:text-6xl font-bold">Series Convergence Cheat Sheet</h1>
<p class="text-xl text-gray-400 mt-2">Your guide to conquering infinite series!</p>
</header>
<!-- NEW: Summary Table Card -->
<div class="card">
<h2>Convergence Tests Summary</h2>
<div class="overflow-x-auto">
<table>
<thead>
<tr>
<th>Test</th>
<th>Conditions & Limit</th>
<th>Conclusion</th>
</tr>
</thead>
<tbody>
<tr>
<td><strong>n-th Term Test for Divergence</strong></td>
<td>\(\lim_{n \to \infty} a_n \neq 0\)</td>
<td>Series <strong>Diverges</strong>. (Inconclusive if limit is 0).</td>
</tr>
<tr>
<td><strong>p-Series</strong><br>\(\sum \frac{1}{n^p}\)</td>
<td>Check the value of \(p\).</td>
<td><strong>Converges</strong> if \(p > 1\).<br><strong>Diverges</strong> if \(p \le 1\).</td>
</tr>
<tr>
<td><strong>Alternating Series Test</strong><br>\(\sum (-1)^n b_n\)</td>
<td>1. \(b_{n+1} \le b_n\) (decreasing)<br>2. \(\lim_{n \to \infty} b_n = 0\)</td>
<td>Series <strong>Converges</strong>. (Check for absolute convergence separately).</td>
</tr>
<tr>
<td><strong>Ratio Test</strong></td>
<td>\(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\)</td>
<td><strong>Abs. Convergent</strong> if \(L < 1\).<br><strong>Divergent</strong> if \(L > 1\).<br><strong>Inconclusive</strong> if \(L = 1\).</td>
</tr>
<tr>
<td><strong>Root Test</strong></td>
<td>\(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\)</td>
<td><strong>Abs. Convergent</strong> if \(L < 1\).<br><strong>Divergent</strong> if \(L > 1\).<br><strong>Inconclusive</strong> if \(L = 1\).</td>
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<!-- The Ratio Test -->
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<h2>The Ratio Test</h2>
<p>This is a powerful test, especially for series involving factorials ($n!$) or exponentials ($k^n$).</p>
<div class="rule-box">
<p>For a series \(\sum a_n\), calculate the limit:</p>
$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
<ul>
<li>If \(L < 1\), the series is <strong>absolutely convergent</strong> (and therefore convergent).</li>
<li>If \(L > 1\) or \(L = \infty\), the series is <strong>divergent</strong>.</li>
<li>If \(L = 1\), the Ratio Test is <strong>inconclusive</strong>. You must use a different test!</li>
</ul>
</div>
<div class="example">
<h3 class="text-2xl text-white mb-2">Example 1: A Convergent Series</h3>
<p>Test the series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\) for absolute convergence.</p>
<p>Here, \(a_n = \frac{2^n}{n!}\) and \(a_{n+1} = \frac{2^{n+1}}{(n+1)!}\).</p>
$$ L = \lim_{n \to \infty} \left| \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} \right| $$
$$ = \lim_{n \to \infty} \left| \frac{2 \cdot 2^n}{(n+1) \cdot n!} \cdot \frac{n!}{2^n} \right| = \lim_{n \to \infty} \left| \frac{2}{n+1} \right| = 0 $$
<p class="conclusion">Since \(L = 0 < 1\), the series is absolutely convergent.</p>
</div>
<div class="example">
<h3 class="text-2xl text-white mb-2">Example 2: A Divergent Series</h3>
<p>Test the series \(\sum_{n=1}^{\infty} \frac{(-3)^n}{n^3}\).</p>
$$ L = \lim_{n \to \infty} \left| \frac{(-3)^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3)^n} \right| $$
$$ = \lim_{n \to \infty} \left| \frac{-3 \cdot n^3}{(n+1)^3} \right| = 3 \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^3 = 3(1)^3 = 3 $$
<p class="conclusion">Since \(L = 3 > 1\), the series diverges.</p>
</div>
<div class="example">
<h3 class="text-2xl text-white mb-2">Example 3: An Inconclusive Result</h3>
<p>Test the series \(\sum_{n=1}^{\infty} \frac{3n}{n+2}\).</p>
$$ L = \lim_{n \to \infty} \left| \frac{3(n+1)}{(n+1)+2} \cdot \frac{n+2}{3n} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n+3} \cdot \frac{n+2}{n} \right| = 1 $$
<p class="conclusion">Since \(L = 1\), the Ratio Test is inconclusive.</p>
<div class="note">
<strong>What now?</strong> Try another test! The Divergence Test is a good choice here.
$$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3n}{n+2} = 3 $$
Since the limit is not 0, the series diverges by the Divergence Test.
</div>
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<!-- Other Important Tests -->
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<h2>Other Key Convergence Tests</h2>
<!-- Divergence Test -->
<div class="rule-box mt-4">
<h3 class="text-xl text-white mb-2">The n-th Term Test for Divergence</h3>
<p>This should be the FIRST test you perform on any series. It can only prove divergence.</p>
$$ \text{If } \lim_{n \to \infty} a_n \neq 0, \text{ the series } \sum a_n \text{ diverges.} $$
<p class="note">If the limit IS 0, this test is inconclusive. The series might converge or diverge.</p>
</div>
<!-- p-Series Test -->
<div class="rule-box mt-4">
<h3 class="text-xl text-white mb-2">p-Series Test</h3>
<p>A quick test for series of the form \(\sum \frac{1}{n^p}\).</p>
$$ \text{The series } \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ is:} $$
<ul>
<li><strong>Convergent</strong> if \(p > 1\).</li>
<li><strong>Divergent</strong> if \(p \le 1\).</li>
</ul>
<p>Example: \(\sum \frac{1}{n^2}\) converges (p=2), but \(\sum \frac{1}{\sqrt{n}}\) diverges (p=1/2).</p>
</div>
<!-- Alternating Series Test -->
<div class="rule-box mt-4">
<h3 class="text-xl text-white mb-2">Alternating Series Test (AST)</h3>
<p>Use for series with alternating signs, like \(\sum (-1)^n b_n\).</p>
<p>The series converges if it meets BOTH conditions:</p>
<ol class="list-decimal list-inside">
<li>\(b_{n+1} \le b_n\) for all n (the terms are decreasing).</li>
<li>\(\lim_{n \to \infty} b_n = 0\).</li>
</ol>
<p class="note">AST only tells you about convergence, not *absolute* convergence. You must test \(\sum |a_n|\) separately to check for that.</p>
</div>
<!-- Root Test -->
<div class="rule-box mt-4">
<h3 class="text-xl text-white mb-2">The Root Test</h3>
<p>A cousin of the Ratio Test. Very useful if the series involves terms raised to the n-th power.</p>
$$ L = \lim_{n \to \infty} \sqrt[n]{|a_n|} $$
<ul>
<li>If \(L < 1\), the series is <strong>absolutely convergent</strong>.</li>
<li>If \(L > 1\) or \(L = \infty\), the series is <strong>divergent</strong>.</li>
<li>If \(L = 1\), the Root Test is <strong>inconclusive</strong>.</li>
</ul>
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