W10 Stepwise

Due date: Thursday 3/19, 11:59pm

Ratio Test and Root Test

01

01

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) ${\displaystyle \sum _{k=0}^{\infty }{\left(\frac{k}{3k+1}\right)}^k}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\frac{e^n}{n!}}$ $$ (c) ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{(2n)!}}$

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Solution

Solutions - 0200-01

(a)

$${\left|{\left(\frac{k}{3k+1}\right)}^k\right|}^{1/k}\gg \gg \frac{k}{3k+1}\stackrel{k\to \infty }{⟶}\frac{1}{3}=L<1$$

Therefore, by the root test the series converges absolutely.

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(b)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|\gg \gg \frac{e^{n+1}}{(n+1)!}\cdot \frac{n!}{e^n}\\ \\ \gg \gg \frac{e}{n+1}\stackrel{n\to \infty }{⟶}0=L<1\end{array}$$

Therefore, by the ratio test the series converges absolutely.

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(c)

$$\begin{array}{c}|a_{n+1}\cdot a_n^{-1}|\gg \gg \frac{1}{(2(n+1))!}\cdot \frac{(2n)!}{1}\\ \\ \gg \gg \frac{1}{(2n+2)(2n+1)}\stackrel{n\to \infty }{⟶}0=L<1\end{array}$$

Therefore, by the ratio test the series converges absolutely.

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Power series: Radius and Interval

02

01

Power series - radius and interval

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

(a) ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n^2{3}^n}}$ $$ (b) ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{n{3}^n}}$ $$ (c) ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{3^n}}$

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Solution

Solutions - 0220-01

(a)

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

Therefore, the radius of convergence is $3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{n^2{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{n^2{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n^2}\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{n^2{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{n^2{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{1}{n^2}\end{array}$$

Both of these series converge, so the final interval of convergence is $I=[-3,+3]$.

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(b)

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

Therefore, $R=3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{n{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{n{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n}{n}\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{n{3}^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{n{3}^n}\\ & \\ & \gg \gg \sum _{n=1}^{\infty }\frac{1}{n}\end{array}$$

The first series converges by the AST. The second diverges ($p\ge 1$).

So the final interval of convergence is $I=[-3,+3)$.

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(c)

$$\begin{array}{c}|a_{n+1}{x}^{n+1}\cdot a_n^{-1}{x}^{-n}|=|\frac{x^{n+1}}{3^{n+1}}\cdot \frac{3^n}{x^n}|\gg \gg \frac{1}{3}|x|\end{array}$$

Therefore, the radius of convergence is $3$ and the preliminary interval is $(-3,+3)$.

Check end points:

$$\begin{array}{cc}x& =-3:\sum _{n=1}^{\infty }\frac{x^n}{3^n}\gg \gg \sum _{n=1}^{\infty }\frac{{(-1)}^n{3}^n}{3^n}\gg \gg \sum _{n=1}^{\infty }{(-1)}^n\\ & \\ x& =+3:\sum _{n=1}^{\infty }\frac{x^n}{3^n}\gg \gg \sum _{n=1}^{\infty }\frac{3^n}{3^n}\gg \gg \sum _{n=1}^{\infty }1\end{array}$$

Both series diverge. So the final interval is $I=(-3,+3)$.

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