Chapter 10.5, Problem 59E

  $a$

To determine

To find: whether the series converges or diverges.

Answer to Problem 59E

Series are: ${\displaystyle \sum }_{n=1}^{\infty }{b}_n$ diverges, ${\displaystyle \sum }_{n=1}^{\infty }{a}_n$ diverges.

Explanation of Solution

Given information:

Given expression is : $A={\displaystyle \sum }_{n=1}^{\infty }\frac{n}{3n^2+1}$

Calculation:

The given series is $A={\displaystyle \sum }_{n=1}^{\infty }\frac{n}{3n^2+1}$

Let $a_n=\frac{n}{3n^2+1}$ and $b_n=\frac{1}{n}$

Then $a_n>0$ and $b_n>0$ for $n\ge 1$

And by the limit comparison test, we have

  $\begin{array}{cc}\underset{n\to \infty }{\lim }\frac{a_n}{b_n}& =\underset{n\to \infty }{\lim }\frac{n^2}{3n^2+1}\\ & =\frac{1}{3}\end{array}$

Since, ${\displaystyle \sum }_{n=1}^{\infty }{b}_n$ diverges, ${\displaystyle \sum }_{n=1}^{\infty }{a}_n$ diverges.

  $b$

To determine

To find : whether the given series is converges or diverges.

Answer to Problem 59E

The given series converges $p=2$ .

Explanation of Solution

Given information:

Given expression is : $A={\displaystyle \sum }_{n=1}^{\infty }\frac{n}{3n^2+1}$ , and $S$ is the series formed by multiplying the $n$ th term in $A$ by the $n$ th term in $n$

  ${\displaystyle \sum }_{n=1}^{\infty }(3/n),$

Calculation:

The other series is

  ${\displaystyle \sum }_{n=1}^{\infty }\frac{3}{n}$

We form a new series $S$ by multiplying the $n$ th term in $A$ by the $n$ th term in the above series. Then

S is given by

  $\begin{array}{cc}S& ={\displaystyle \sum }_{n=1}^{\infty }\frac{n}{3n^2+1}\times \frac{3}{n}\\ & ={\displaystyle \sum }_{n=1}^{\infty }\frac{3}{3n^2+1}\end{array}$

This series $S$ converges by the direct comparison test, since $\frac{3}{3n^2+1}<\frac{1}{n^2}$ and ${\displaystyle \sum }_{n=1}^{\infty }\frac{1}{n^2}$ converges by the $p$ -test with $p=2$ .