Chapter 10.5, Problem 19E

To determine

To find: whether the series converges or diverges.

Answer to Problem 19E

Hence, it has not satisfied third condition of test. Thus, series diverges

Explanation of Solution

Given information:

Given expression: ${\displaystyle \sum }_{n=1}^{\infty }{(-1)}^{n+1}\frac{{10}^n}{n^{10}}$

Calculation:

The Alternating Series Test (Leibniz's Theorem):

The series

  ${\displaystyle \sum }_{n=1}^{\infty }{(-1)}^{n+1}{u}_n=u_1-u_2+u_3-u_4+\dots \dots \mathrm{..}$

Converges if all the three of the following conditions are satisfied:

1. Each $u_n$ is positive.

2. $u_n\ge u_{n+1}$ for all $n\ge N$ for some integer $N$

3. $\underset{n\to \infty }{\lim }{u}_n\to 0$

Given series ${\displaystyle \sum }_{n=1}^{\infty }{(-1)}^{n+1}\frac{{10}^n}{n^{10}}$

Here,

  $u_n=\frac{{10}^n}{n^{10}}\text{for}n\ge 1$

Hence, $u_n$ is positive.

Since, $\frac{{10}^n}{n^{10}}$ is decreasing for $n\ge 1$

Thus, $u_n\ge u_{n+1}$

Now,

  $\underset{n\to \infty }{\lim }{u}_n=\underset{n\to \infty }{\lim }\frac{{10}^n}{n^{10}}$

Apply L' Hospital's rule

  $\begin{array}{l}=\underset{n\to \infty }{\lim }\frac{{(\ln 10)}^{10}{10}^n}{10!}\hfill \\ \to \infty \hfill \end{array}$

Since, it has not satisfied third condition of test. Thus, series diverges