Chapter 9.5, Problem 29E

To determine

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

Answer to Problem 29E

The series converges conditionally.

Explanation of Solution

Given information:

  $\left|a_n\right|=1/(1+\sqrt{n})$

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

Since the general term of the series $\left|a_n\right|$ behaves like $1/\sqrt{n}$ for large $n$ , should compare the general term of the series to terms of the series $1/\sqrt{n}$ and use the limit comparison test.

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

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

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

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

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

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

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

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

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

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

Converges conditionally.

Therefore, the series converges conditionally.