Chapter 9.5, Problem 32E

To determine

To calculate: The series converges absolutely, converges conditionally, or diverges.

Answer to Problem 32E

The series converges conditionally.

Explanation of Solution

Given information:

  $\left|a_n\right|={(-1)}^n/(\sqrt{n}+\sqrt{n+1})$

Calculation:

Let examine whether the series converges absolutely. If it does not, let use other tests in order to determine conditional convergence.

First examine the series $\left|a_n\right|={(-1)}^n/(\sqrt{n}+\sqrt{n+1})$ . If this series converges, then the series converges absolutely.

Can give a lower bound for the series $\left|a_n\right|$

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n}+\sqrt{n+1}}}\\ \ge {\displaystyle \sum _{n=1}^{\infty }\frac{1}{2\sqrt{n+1}}}\\ {\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}=\frac{1}{2}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{\sqrt{n+1}}}\end{array}$

Determine whether the series $1/(\sqrt{n+1})$ converges or diverges. Since the general term of the series behaves like $1/\sqrt{n}$ for large $n$ , can use the limit comparison test.

Let $a_n=1/\sqrt{n+1}$ and $b_n=1/\sqrt{n}$ , therefore based on the limit comparison test

  $\begin{array}{c}\underset{n\to \infty }{\lim }\frac{a_n}{b_n}=\underset{n\to \infty }{\lim }\frac{\frac{1}{\sqrt{n+1}}}{\frac{1}{\sqrt{n}}}\\ =\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n+1}}\cdot \frac{\sqrt{n}}{1}\\ =\underset{n\to \infty }{\lim }\frac{\sqrt{n}}{\sqrt{n+1}}=\\ \underset{n\to \infty }{\lim }\sqrt{\frac{n}{n+1}}\\ \underset{n\to \infty }{\lim }\frac{a_n}{b_n}=1\end{array}$

By using the $p$ -series test obtain that ${\displaystyle \sum 1}/\sqrt{n}={\displaystyle \sum 1}/n^{1/2}$ diverges, because $p=1/2<1$

Based on the previous two steps, by using the direct comparison test $\left|a_n\right|$ diverges, hence the original series does not converge absolutely.

Let check whether the series converges conditionally by using the alternating seriestest.

Since $\sqrt{n}$ is positive for every $n\ge 1$ , also $u_n=1/(\sqrt{n}+\sqrt{n+1})$ is positive, i.e. the first condition is satisfied.

Since $\sqrt{n}$ is increasing, hence $u_n=1/(\sqrt{n}+\sqrt{n+1})$ is decreasing for every $n$ , thus also the second condition is satisfied.

Finally, $\sqrt{n}$ grows without bound as $n$ approaches infinity, therefore $u_n=1/(\sqrt{n}+\sqrt{n+1})$ approaches zero, i.e. also the third condition is satisfied.

Since $u_n$ satisfies all the three conditions of the the alternating series test, based on the test the series

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^n}{\sqrt{n}+\sqrt{n+1}}}$

converges conditionally.

Therefore, the it conditionally converges.