Chapter 10.5, Problem 18E

To determine

To find: whether the series converges or diverges.

Answer to Problem 18E

Hence, it satisfies all three condition of test. Thus, series converges.

Explanation of Solution

Given information:

Given expression: ${\displaystyle \sum }_{n=2}^{\infty }{(-1)}^{n+1}\frac{1}{\ln n}$

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 ${\displaystyle \sum }_{n=2}^{\infty }{(-1)}^{n+1}\frac{1}{\ln n}$

Here,

  $u_n=\frac{1}{\ln n}\text{ for }n\ge 2$

Hence, $u_n$ is positive.

Since, $\frac{1}{\ln n}$ is decreasing for $n\ge 2$

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

Now,

  $\begin{array}{cc}\underset{n\to \infty }{\lim }{u}_n& =\underset{n\to \infty }{\lim }\frac{1}{\ln n}\\ & \to 0\end{array}$

Since, it satisfies all three condition of test. Thus, series converges.