Chapter 10, Problem 43RE

To determine

To find: If the given series converges absolutely, converges conditionally or diverges.

Answer to Problem 43RE

Series 3

  ${\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n^3}}$ converges absolutely.

Explanation of Solution

Given:

The given series is, ${\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n^3}}$

Calculation:

By direct comparison test, consider the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$

Since using p -test series test, ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ converges if p >1, and diverges if p $\le$ 1

Series is in the form of ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ , where p=2 . Thus, series converges, since p >1

Since, the limit is positive and ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}$ converges.

Since, $0\le {\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n^3}}<{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}\text{and}{\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2}}}$ converges.

Thus, ${\displaystyle \sum _{n=1}^{\infty }\frac{\ln n}{n^3}}$ converges absolutely.