Chapter 10.5, Problem 62E

  $a$

To determine

To show: the series converges for $p>1$ .

Answer to Problem 62E

The series converges by the Direct Comparison Test, since $\frac{1}{n^p\ln n}<\frac{1}{n^p}$ for $n\ge 3$ and ${\displaystyle \sum }_{n=3}^{\infty }\frac{1}{n^p}$ converges as a $p$ = series when $p>1$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^p\ln n},p\ge 0$

Calculation:

The given series is ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^p\ln n},p\ge 0$

The series converges by the Direct Comparison Test, since $\frac{1}{n^p\ln n}<\frac{1}{n^p}$ for $n\ge 3$ and ${\displaystyle \sum }_{n=3}^{\infty }\frac{1}{n^p}$ converges as a $p$ =series when $p>1$

  $b$

To determine

To find : whether the series converges or diverges for $p=1$ .

Answer to Problem 62E

The answer for the expression is : $=\infty$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^p\ln n},p\ge 0$

Calculation:

For $p=1$ the series is

  ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^p\ln n}={\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n\ln n}$

This diverges by the Integral Test, since

  $\begin{array}{cc}{{\displaystyle \int }}_2^{\infty }\frac{1}{x\ln x}dx& =\underset{b\to \infty }{\lim }{[\ln (\ln x)]}_2^b\\ & =\infty \end{array}$

  $c$

To determine

To show: the series diverges for $0\le p\le 1$ .

Answer to Problem 62E

The answer for the expression is : ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n\ln n}$

Explanation of Solution

Given information:

Given expression is : ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^p\ln n},p\ge 0$

Calculation:

For $0\le p\le 1,$ we have $\frac{1}{n^{\prime }\ln n}>\frac{1}{n\ln n},$ so ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n^{\prime }\ln n}$ diverges by the Direct Comparison

Test with ${\displaystyle \sum }_{n=2}^{\infty }\frac{1}{n\ln n}$ from part (b).