W10 Regular

Due date: Sunday 3/22, 11:59pm

Ratio Test and Root Test

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Ratio and root tests

Apply the ratio test or the root test to determine whether each of the following series is absolutely convergent, conditionally convergent, or divergent.

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-2)}^n}{n^{100}}}$ $$ (b) ${\displaystyle \sum _{n=0}^{\infty }{\left(\frac{5n}{10n+4}\right)}^n}$ $$ (c) ${\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{3^n}}$

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Solution

Solutions - 0200-02

(a)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|\gg \gg \frac{2^{n+1}}{{(n+1)}^{100}}\cdot \frac{n^{100}}{2^n}\\ \\ \gg \gg 2{\left(\frac{n}{n+1}\right)}^{100}\stackrel{n\to \infty }{⟶}2=L>1\end{array}$$

Since $L>1$, the ratio test says that the series diverges.

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(b)

$$\sqrt[n]{|a_n|}=\frac{5n}{10n+4}\stackrel{n\to \infty }{⟶}\frac{5}{10}=L<1$$

Since $L<1$, the root test says that the series converges.

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(c)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=\frac{\sqrt{n+1}}{3^{n+1}}\cdot \frac{3^n}{\sqrt{n}}\\ \\ \gg \gg \frac{1}{3}\sqrt{\frac{n+1}{n}}\stackrel{n\to \infty }{⟶}\frac{1}{3}=L<1\end{array}$$

Since $L<1$, the ratio test says that the series converges.

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Series tests: strategy tips

02 $★$

01

Various limits, Part I

Find the limits. You may use $+\infty$ or $-\infty$ or $\mathrm{DNE}$ as appropriate. Braces indicate sequences.

• C = Convergent
• AC = Absolutely Convergent
• CC = Conditionally Convergent
• D = Divergent

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{1}{n+2}}$
${\displaystyle \frac{n}{n+2}}$
${\displaystyle \frac{1}{n^2+2}}$
${\displaystyle \frac{4}{2^n}}$
${\displaystyle \frac{4n}{2^n}}$

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Solution

Solutions - 0210-01

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{1}{n+2}}$	0	$C$	0	$C$	$D$	$CC$
${\displaystyle \frac{n}{n+2}}$	1	$C$	$DNE$	$D$	$D$	$D$
${\displaystyle \frac{1}{n^2+2}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{4}{2^n}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{4n}{2^n}}$	0	$C$	0	$C$	$AC$	$AC$

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03 $★$

02

Various limits, Part II

Find the limits. You may use $+\infty$ or $-\infty$ or $\mathrm{DNE}$ as appropriate. Braces indicate sequences.

• C = Convergent
• AC = Absolutely Convergent
• CC = Conditionally Convergent
• D = Divergent

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{4n!}{2^n}}$
${\displaystyle \frac{(n+2)3^n}{n!}}$
${\displaystyle \frac{4^n}{{(3n)}^n}}$
${\displaystyle \frac{1}{(2n+1)!}}$

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Solution

Solutions - 0210-02

$a_n$	${\lim }_{n\to \infty }{a}_n$	$\left\{a_n\right\}$ C or D	${\lim }_{n\to \infty }{(-1)}^n{a}_n$	$\left\{{(-1)}^n{a}_n\right\}$ C or D	$\sum a_n$ AC, CC, or D	$\sum {(-1)}^n{a}_n$ AC, CC, or D
${\displaystyle \frac{4n!}{2^n}}$	$\infty$	$D$	$DNE$	$D$	$D$	$D$
${\displaystyle \frac{(n+2)3^n}{n!}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{4^n}{{(3n)}^n}}$	0	$C$	0	$C$	$AC$	$AC$
${\displaystyle \frac{1}{(2n+1)!}}$	0	$C$	0	$C$	$AC$	$AC$

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Power series: Radius and Interval

04 $★$

02

Power series - radius and interval

Find the radius and interval of convergence for these power series:

(a) ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(x+3)}^n}{n!}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\frac{{(x-7)}^n}{n}}$ $$ (c) ${\displaystyle \sum _{n=12}^{\infty }{n}^n{(x-2)}^n}$

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Solution

Solutions - 0220-02

(a)

$$\begin{array}{c}|\frac{{(x+3)}^{n+1}}{(n+1)!}\cdot \frac{n!}{{(x+3)}^n}|\gg \gg \frac{1}{n+1}|x+3|\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore $R=\infty$ and $I=(-\infty ,+\infty )$.

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(b)

$$|\frac{{(x-7)}^{n+1}}{n+1}\cdot \frac{n}{{(x-7)}^n}|\gg \gg \frac{n}{n+1}|x-7|\stackrel{n\to \infty }{⟶}0$$

Therefore $R=1$ and the preliminary interval is $(\mathrm{6,8})$.

At $x=6$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(-1)}^n}{n}=\sum _{n=0}^{\infty }\frac{1}{n}}$. This diverges ($p=1$).

At $x=8$ we have ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{{(+1)}^n}{n}=\sum _{n=0}^{\infty }\frac{{(-1)}^n}{n}}$. This converges by the AST.

Therefore, the final interval of convergence is $I=(\mathrm{6,8}]$.

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(c)

$$\left|\frac{{(n+1)}^{n+1}{(x-2)}^{n+1}}{n^n{(x-2)}^n}\right|\gg \gg (n+1){(1+\frac{1}{n})}^n|x-2|$$

Observe that $1+\frac{1}{n}>1$ so ${\lim }_{n\to \infty }{(1+\frac{1}{n})}^n\ge 1$. Assume $x\ne 2$. Then:

$$(n+1){(1+\frac{1}{n})}^n|x-2|\stackrel{n\to \infty }{⟶}\infty$$

If $x=2$, then of course the series is $\sum 0$ and converges to $0$.

Therefore $R=0$ and $I=\{2\}$.

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05

03

Power series - radius and interval

Find the radius and interval of convergence for these power series:

(a) ${\displaystyle \sum _{n=0}^{\infty }\frac{{(x-8)}^n}{n^4+1}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{3\cdot 7\cdot 11\cdots (4n-1)}}$

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Solution

Solutions - 0220-03

(a)

Apply the ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{{(x-8)}^{n+1}}{{(n+1)}^4+1}\cdot \frac{n^4+1}{{(x-8)}^n}|\\ \\ \gg \gg \frac{n^4+1}{{(n+1)}^4+1}|x-8|\\ \\ \gg \gg \frac{n^4+1}{n^4+4n^3+6n^2+4n+1}|x-8|\stackrel{n\to \infty }{⟶}|x-8|\end{array}$$

Therefore $R=1$ and the preliminary interval is $(\mathrm{7,9})$.

Check endpoints:

At $x=7$, we have ${\displaystyle \sum _{n=0}^{\infty }\frac{{(-1)}^n}{n^4+1}}$, which converges absolutely.

At $x=9$, we have ${\displaystyle \sum _{n=0}^{\infty }\frac{1}{n^4+1}}$, which converges by the DCT, comparing with ${\displaystyle \sum \frac{1}{n^4}}$.

Therefore, the final interval of convergence is $I=[\mathrm{7,9}]$.

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(b)

Apply the ratio test:

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|=|\frac{x^{n+1}}{3\cdot 7\cdot 11\cdots (4n-1)(4n+3)}\cdot \frac{3\cdot 7\cdot 11\cdots (4n-1)}{x^n}|\\ \\ \gg \gg \frac{1}{4n+3}|x|\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore, $R=\infty$ and the interval of convergence is $I=(-\infty ,+\infty )$.

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