Calc II Study Guide Series

Step 1: Check if the series is divergent:

• $\sum_{n=1}^{\infty}a_{n}$ if $lim_{n\rightarrow\infty}a_{n}\ne0$

Step 2: Is the series one of the popular series (Geometric series, p-series, alternating series)?

Geometric series: $\sum_{n=0}^\infty a r^n$

• Converges if -1 < r < 1

P-series: $\sum_{n=1}^\infty \frac 1{n^P}$

• Converges if p > 1

Alternating series: $\sum_{n=1}^\infty (-1)^n a_n$

• Converges if a_n > a_{n+1} > 0 and $lim_{n \rightarrow \infty} a_n = 0$

Step 3: Check if you can use the ordinary comparison test or the limit comparison test

The Comparison test

• Suppose $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are positive terms

  • If $\sum_{n=1}^\infty b_n$ converges and $a_n \le \ b_n$ for all n, then $\sum_{n=1}^\infty$ converges

  • If $\sum_{n=1}^\infty b_n$ diverges and $a_n \ge \ b_n$ for all n, then $\sum_{n=1}^\infty a_n$ diverges

Limit Comparison Test

• If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ are positive-term series, and if $lim_{n \rightarrow \infty} \frac {a_n}{b_n} = K$ where K > 0 and finite, then either both series converge or both diverge

Step 4: Check if you can use the ratio test. For a series $\sum a_n$, let $L = lim_{n \rightarrow \infty} |\frac {a_{n+1}}{a_n}|$

• If L < 1, the series is absolutely convergent and thus convergent

• If L > 1, the series is divergent

• If $L = 1$ the test fails; no conclusion regarding convergence/divergence

Step 5: Check if you can use the absolute convergence test. (if the question asks about ABSOLUTE CONVERGENCE)

• If the series $\sum |a_n|$ is convergent, then $\sum a_n$ is absolutely convergent

Step 6: Check if you can use the integral test

• Let f be a function that is positive-valued, continuous and decreasing on $[1, \infty)$, and $a_n = f(n)$ for all integers n. Then

  • $\int_1 ^\infty f(x) dx$

  • $\sum_{n}^\infty f(n)$

• either both converge or diverges

Steps for determining whether the series $\sum_{n=1}^\infty a_n$ is convergent or not

1. Check if the series is divergent (Use the divergence test)

2. Is the series one of the popular series (Geometric series, p-series, alternating series)?

3. Check if you can use the ordinary comparison test or the limit comparison test.

4. Check if you can use the ratio test

5. Check if you can use the absolute convergence test (if the question asks about absolute convergence)

6. Check if you can use the integral test.

a) $\frac {3n-1}{6n+2}$

$\lim_{n\to \infty} \frac {3n-1}{6n+2} = \frac 36 = \frac 12$

$\frac 12 \le 1$

Convergent to $\frac 12$

b) $\frac {\cos(n)}{n}$

$a_n = \frac {\cos n}n$

$|\cos n| \le 1 \Rightarrow |\frac {\cos n}n| \le \frac 1n \overrightarrow{n\to \infty} 0$

Squeeze Theorem using $-\frac 1n \le \frac {\cos n} n \le \frac 1n \rightarrow 0. \Rightarrow$ Convergent to 0

c) $(-\frac 23) ^n$

1)^n a_n$</p><p>$a_1 = (-\frac 23)^1 = -\frac 23$</p><p>$a_2 = (-\frac 23)² = \frac 49$</p><p>$a_3 = (-\frac 23)³ = -\frac 8{27}$</p><p>$r = -\frac 23$</p><p>$\\|r| = \frac 23 < 1 \Rightarrow r^n \rightarrow 0$</p><p></p><p>d)$\frac {5^n}{3^{n+1}}$</p><p></p><p>$\frac 13(\frac 53)^n$</p><p>$r = \frac 53$</p><p>$|r| = \frac 53 > 1 \Rightarrow r^n \rightarrow \infty$</p><p></p><p>e)$\frac{\sin²n}{2^n}$</p><p></p><p>$|\sin² n| \le 1 \Rightarrow |\frac {\sin² n}{2^n}| \le \frac 1{2^n} \overrightarrow {n \rightarrow \infty}0$</p><p></p><ol start="8"><li><p>For each of the following sequences, find a formula for the <em>n</em>th term,$a_n, assuming that the indicated pattern continues.

• {\frac 12, \frac 23, \frac 34, \frac 45, …}

a_1 = \frac 12$</p><p>$a_2 = \frac 23$</p><p>$a_3 = \frac 34$</p><p>$a_4 = \frac 45$</p><p></p><p>$a_n = \frac n{n+1}

• {\frac 12, -\frac 45,\frac 98, …}

a_1 = \frac 12$</p><p>$a_2 -\frac 45$</p><p>$a_3 = \frac 98$</p><p></p><p>$a_n = (-1)^{n+1}(\frac {n²}{3n - 1})$</p><p></p><p><strong><u>9. </u></strong>Determine if the following series converges or diverges. If it converges, find its sum.</p><p>$\frac 1{3\cdot 4} + \frac 1{4 \cdot 5} + \frac 1{5 \cdot 6} + … + \frac 1{(n+2)(n+3)} + …$</p><p></p><p>$a_n = \frac 1{(n+2)(n+3)} = \frac 1{n+2} - \frac 1{n+3}$</p><p></p><p>Consider the <em>N</em>th partial sum:</p><p>$S_N = \sum_{n=1}^N (\frac 1{n+2} - \frac 1{n+3}) =$</p><p>$\frac 13- \frac 14 + \frac 14 - \frac 15 + \frac 15 - \frac 16$</p><p></p><p>All intermediate terms cancel, leaving:</p><p>$S_N = \frac 13 - \frac 1{N+3}$</p><p>As$N\to \infty$,$\sum_{n=1}^\infty \frac 1{(n+2)(n+3)} \Rightarrow \frac 13 - \frac 1{N+3} = \frac 13 + 0 = \frac 13$</p><p>Since$\frac 13 \le 1$, this series converges to$\frac 13$</p><p></p><p><strong>10.</strong> Do the following terms converge or diverge:</p><p></p><p style="text-align: center;">a)$\sum_{n=1}^\infty (\frac 32)^n$</p><p style="text-align: center;"></p><p>$a_n = (\frac 32)^n$; This is a geometric series</p><p>$r = \frac 32$</p><p>Since$\frac 32 > 1$, this series diverges</p><p></p><p style="text-align: center;">b)$\sum_{n=1}^\infty 3^n\cdot 5^{-n}$</p><p style="text-align: center;"></p><p>$a_n = 3^n \cdot 5^{-n} \Rightarrow 3^n \cdot \frac 1{5^n} = \frac {3^n}{5^n} = (\frac 35)^n$</p><p>$r = 35$</p><p>Since$\frac 35 < 1$, this series converges to$\frac r{1-r} = \frac {\frac 35}{1 -\frac 35} = \frac 32$</p><p>This is geometric with ratio$r = \frac 35$and$|r| < 1$. Therefore it converges. If needed, the sum from$n=1$is$\frac 32$</p><p></p><p></p><p style="text-align: center;">c)$\sum_{n=1}^\infty (-1)^n \frac 1{3n+1}$</p><p></p><p>This is a Alternating series. which converges if:</p><ul><li><p>$a_n > a_{n+1} > a_{n+2} + …$</p></li><li><p>$\lim_{n\to \infty} a_n= 0$</p></li></ul><p>$a_n = \frac 1{3n+1}$</p><p></p><p>Testing for part 1:</p><ul><li><p>$a_1 = \frac 1{3(1) + 1} = \frac 14$</p></li><li><p>$a_2 = \frac 1{3(2) + 1} = \frac 17$</p></li><li><p>$a_3 = \frac 1{3(3) + 1} = \frac 1{10}$</p></li></ul><p></p><p>$a_1 > a_2 > a_3 > … \Rightarrow$True</p><p></p><p>Testing for part 2:</p><p>$\lim_{n\to\infty} \frac 1{3n+1} \rightarrow \frac 1\infty = 0$</p><p>$lim_{n\to\infty} a_n = 0 \Rightarrow$True</p><p></p><p>By the alternating series test, this series <strong><u>converges</u></strong></p><p></p><p><strong><u>11.</u> </strong>For the following items<u>,</u> use the <u>comparison test</u> or the <u>limit comparison test</u> to determine whether the series is convergent or divergent.</p><p></p><p style="text-align: justify;">The Comparison test</p><ul><li><p style="text-align: justify;">Suppose$\sum_{n=1}^\infty a_n$and$\sum_{n=1}^\infty b_n$are positive terms</p><ul><li><p style="text-align: justify;">If$\sum_{n=1}^\infty b_n$converges and$a_n \le \ b_n$for all n, then$\sum_{n=1}^\infty$converges</p></li><li><p style="text-align: justify;">If$\sum_{n=1}^\infty b_n$diverges and$a_n \ge \ b_n$for all n, then$\sum_{n=1}^\infty a_n$diverges</p></li></ul></li></ul><p style="text-align: justify;">Limit Comparison Test</p><ul><li><p style="text-align: justify;">If$\sum_{n=1}^\infty a_n$and$\sum_{n=1}^\infty b_n$are positive-term series, and if$lim_{n \rightarrow \infty} \frac {a_n}{b_n} = K where K > 0 and finite, then either both series converge or both diverge

a)\sum_{n=5}^\infty \frac 1{\sqrt n -2}$</p><p></p><p>$a_n = \frac 1{\sqrt n - 2}$</p><p>$b_n = \frac 1{\sqrt n}$</p><p></p><p>$\frac 1{\sqrt n} = \frac 1{n^{\frac 12}}$, which is a p-series where$p = \frac 12. Since p < 1, the series diverges.

The Limit Comparison Test:

To check if a_n$also diverges, we need to use the Limit Comparison test.</p><p>$\frac {a_n}{b_n} = \frac {\frac {1}{\sqrt n - 2}}{\frac 1{\sqrt n}} = \frac {\sqrt n}{\sqrt n - 2} \Rightarrow \frac {\sqrt n}{\sqrt n - 2} \cdot \frac {\frac 1{\sqrt n}}{\frac 1 {\sqrt n}} = \frac 1{1 - \frac 2{\sqrt n}} = 1$</p><p></p><p>The limit is a finite positive constant (1), so therefore, the limit of$a_n$will have the same behavior and therefore, divergent</p><p></p><p style="text-align: center;">b)$\sum_{n=1}^\infty \frac 1{4^n +3}$</p><p style="text-align: left;"></p><p style="text-align: left;">$a_n = \frac 1 {4^n + 3}$. Since$a_n$approaches 0 as$n\to\infty$,$\sum_{n=1}^\infty \frac 1{4^n +3}$converges </p><p style="text-align: center;"></p><p style="text-align: left;"><strong><u>12.</u></strong> </p><p style="text-align: center;">$\sum_{n=1}^\infty ne^{-n²}$</p><p style="text-align: left;"></p><p style="text-align: left;">$a_n = n e^{-n²}$</p><p style="text-align: left;"></p><p style="text-align: left;">$ne^{-n²} = \frac n{e^{n²}}$</p><p style="text-align: left;"></p><p style="text-align: left;">$a_1 = \frac 1e$</p><p style="text-align: left;">$a_2 = \frac 2{e^4}$</p><p style="text-align: left;">$a_3 = \frac 3 {e^9}

As \lim_{n\to\infty}, a_n = 0$$

Therefore, this series converges