Chapter 10.5, Problem 21E

To determine

Whether the series converges or diverges.

Answer to Problem 21E

The series is not either convergent or divergent, it is oscillating.

Explanation of Solution

Given information:

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

To check whether the series converges or diverges:

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

The sum is either $\frac{1}{2}$ or $-\frac{1}{2}$ .

This implies

The sum is oscillating.

Thus,

The series is also oscillating.

Therefore,

The series is not either convergent or divergent.