Chapter 10.5, Problem 41E

a.

To determine

To find:The interval of convergence of given series.

Answer to Problem 41E

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ converges in interval $-3\le x\le 3$ or $x\in \left[-3,3\right]$ .

Explanation of Solution

Given:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ .

Formula/ concept used:

Ratio Test: Suppose ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ be a given series and let $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ , then if $L<1$ , the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ is convergent (absolutely convergent) and if $L>1$ the series is divergent.

Calculations:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$.........(1)

Now, $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\underset{n\to \infty }{\lim }\left|\frac{\frac{x^{n+1}}{(n+1)\sqrt{n+1}{3}^{n+1}}}{\frac{x^n}{n\sqrt{n}{3}^n}}\right|$

  $L=\underset{n\to \infty }{\lim }\left|\frac{x^{n+1}}{(n+1)\sqrt{n+1}{3}^{n+1}}\frac{n\sqrt{n}{3}^n}{x^n}\right|$

  $L=\underset{n\to \infty }{\lim }\left|\frac{x}{(1+\frac{1}{n})\sqrt{1+\frac{1}{n}}}\frac{1}{3}\right|$

  $L=\left|\frac{x}{\left(1+0\right)\sqrt{1+0}}\frac{1}{3}\right|=\left|\frac{x}{3}\right|$$\because \underset{n\to \infty }{\lim }\frac{a}{n}=0$

The series (1) is convergent if $L<1\Rightarrow \left|\frac{x}{3}\right|<1$

  $\Rightarrow -3<x<3$

Now, for $x=3$ , ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}={\displaystyle \sum _{n\text{=0}}^{\infty }\frac{3^n}{n\sqrt{n}{3}^n}}={\displaystyle \sum _{n\text{=0}}^{\infty }\frac{1}{n\sqrt{n}}}$ that is convergent series.

Hence, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ converges in interval $-3\le x\le 3$ or $x\in \left[-3,3\right]$ .

Conclusion:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ converges in interval $-3\le x\le 3$ or $x\in \left[-3,3\right]$ .

b.

To determine

To find:The values of x for which the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ is absolutely convergent.

Answer to Problem 41E

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ is absolutely convergent in interval $-3\le x\le 3$ .

Explanation of Solution

Given:

Theseries ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ in part (a).

Concept used:

Suppose ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ be a given series and let $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ , then if $L<1$ , the series ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ is convergent (absolutely convergent) and if $L>1$ the series is divergent.

Calculations:

The given series is ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ .

As shown in part (a) $L=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=\left|x\right|<1$ in interval $-3\le x\le 3$ .

Therefore, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ is absolutely convergent in interval $-3\le x\le 3$ .

Conclusion:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ is absolutely convergent in interval $-3\le x\le 3$ .

c.

To determine

To find: The value of x for which the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ is conditionally convergent.

Answer to Problem 41E

There is no interval in which the series converge conditionally.

Explanation of Solution

Given:

The series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ in part (a).

Concept used:

The series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ is conditionally convergent if the series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ converges but ${\displaystyle \sum _{n\text{=}1}^{\infty }\left|a_n\right|}$ diverges.

If the original series ${\displaystyle \sum _{n\text{=}1}^{\infty }{a}_n}$ is absolutely convergent then series does not converge conditionally.

If a series is alternating and not absolutely convergent, we check for conditional convergence using the alternating series test.

Calculations:

As shown in part (a) and (b) the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}$ and $\left|{\displaystyle \sum _{n\text{=0}}^{\infty }\frac{x^n}{n\sqrt{n}{3}^n}}\right|$ both converge in the interval $-3<x<3$ .

Since, the series ${\displaystyle \sum _{n\text{=0}}^{\infty }\frac{n{x}^n}{n+2}}$ absolutely convergentin interval, $-3<x<3$ , therefore there is no interval in which the series converge conditionally.

Conclusion:

There is no interval in which the series converge conditionally.