Chapter 9.5, Problem 49E

a.

To determine

To find: The interval of convergence of the series.

Answer to Problem 49E

The interval of convergence of the series is $-1-\pi <x<1-\pi$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+\pi \right)}^n}{\sqrt{n}}}$

Calculation:

By applying the Ratio Test and looking at the interval's endpoints, one can determine the interval where the series will converge.

Taking the absolute value of the terms in the series and looking at the following limit, only using the Ratio Test for series with non-negative terms.

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{\left|\frac{{\left(x+\pi \right)}^{n+1}}{\sqrt{n+1}}\right|}{\left|\frac{{\left(x+\pi \right)}^n}{\sqrt{n}}\right|}=\underset{n\to \infty }{\lim }\left(\frac{{\left|x+n\right|}^{n+1}}{\sqrt{n+1}}.\frac{\sqrt{n}}{{\left|x+n\right|}^n}\right)\\ =\underset{n\to \infty }{\lim }\left(\sqrt{\frac{n}{n+1}}\cdot \left|x+\pi \right|\right)\\ =\left|x-\pi \right|\end{array}$

The series converges absolutely based on the previous step if $\left|x-\pi \right|<1$ i.e., $-1<x-\pi <1$ , from which the interval of convergence is given by,

  $-1-\pi <x<1-\pi$

The left endpoint of the interval is given by,

  $\left(-1-\pi ,1-\pi \right)$

Where,

  $x=-1-\pi$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1-\pi +\pi \right)}^n}{\sqrt{n}}}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{1}{n^{1/2}}$

The series does not absolutely converge at the left endpoint since the $p$ -series test indicates that it diverges; even though it is absolute convergent if ${\displaystyle \sum \left|a_n\right|={\displaystyle \sum 1/n^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$ converges.

Analyze if the series conditionally converges at the left endpoint. The series at the left endpoint according to the preceding step is

  ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{1}{n^{1/2}}$

The series converges at the left endpoint according to the Alternating Series Test because $1/n^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ is positive, decreasing, and equal to one for every $n$ , and it gets closer to zero as n gets closer to infinity.

The right endpoint of the interval is given by,

  $\left(-1-\pi ,1-\pi \right)$

Where,

  $x=1-\pi$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(1-\pi +\pi \right)}^n}{\sqrt{n}}}={\displaystyle \sum _{n=0}^{\infty }\frac{1}{n^{1/2}}}$

The series does not converge at the proper endpoint since it has only positive terms, is absolute convergent if it does, never converges conditionally, and diverges according to the $p$ -series test.

From the above steps it is known that,

The interval of convergence of the series is,

  $-1-\pi <x<1-\pi$

Therefore, the interval of convergence of the series is $-1-\pi <x<1-\pi$ .

b.

To determine

To find: For what value of $x$ does the series converge absolutely.

Answer to Problem 49E

The value of $x$ for which the series converges absolutely is $-1-\pi <x<1-\pi$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+\pi \right)}^n}{\sqrt{n}}}$

Calculation:

From part (a) it is known that,

The interval on which the series converges absolutely is given by,

  $\left(-1-\pi ,1-\pi \right)$

Using the Ratio test,

The series diverges if,

  $x\notin \left[-1-\pi ,1-\pi \right]$

Then, the series converges absolutely on $\left(-1-\pi ,1-\pi \right)$ .

Therefore, the value of $x$ for which the series converges absolutely is $-1-\pi <x<1-\pi$ .

c.

To determine

To find: For what value of $x$ does the series converge conditionally.

Answer to Problem 49E

The value of $x$ for which the series converges conditionally is $x=-1-\pi$ .

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(x+\pi \right)}^n}{\sqrt{n}}}$

Calculation:

Identifying the intervals in parts (a) and (b) on which the series converges and on which it converges absolutely.

These findings can be used to determine where the series conditionally converges by looking at the conditional convergence of the endpoints.

From part (a) it is known that,

The interval on which the series converges is given by,

  $x=-1-\pi$

But from part (a) and part (b) it is found that it does not converge absolutely at this point.

Then, the series converges conditionally on $x=-1-\pi$ .

Therefore, the value of $x$ for which the series converges conditionally is $x=-1-\pi$ .