Chapter 9.5, Problem 16E

To determine

To find: Whether the given series converges or diverges.

Answer to Problem 16E

The given series is convergent.

Explanation of Solution

Given information:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{5n^3-3n}{n^2\left(n+2\right)\left(n^2+5\right)}}$

Calculation:

The direct comparison test and the $p$ -series test can be used to identify if the series is converging or diverging.

To get an upper bound for the given series, first rewrite the general term.

  $\begin{array}{c}\frac{5n^3-n}{n^2\left(n+2\right)\left(n^2+5\right)}=\frac{5n^2-1}{n\left(n+2\right)\left(n^2+5\right)}\\ =\frac{5n^2-3}{n^4+2n^3+5n^2+10n}\end{array}$

Thus, providing an upper bound for the series based on the preceding step:

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\frac{5n^3-n}{n^2\left(n+2\right)\left(n^2+5\right)}}={\displaystyle \sum _{n=1}^{\infty }\frac{5n^2-3}{n^4+2n^3+5n^2+10n}}\\ \le {\displaystyle \sum _{n=1}^{\infty }\frac{5n^2-3}{n^4}}\\ ={\displaystyle \sum _{n=1}^{\infty }\frac{5}{n^2}}\end{array}$

Since, $n>0$

Using the $p$ -series test, the series ${\displaystyle \sum _{}^{}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$n^2$}\right.}$ converges and the series ${\displaystyle \sum _{}^{}\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$n^2$}\right.}$ also converges.

  $\begin{array}{c}{\displaystyle \sum _{}^{}\frac{5}{n^2}}=5{\displaystyle \sum \frac{1}{n^2}}\\ <\infty \end{array}$

Using the direct comparison test, determining the given series converges based on the previous steps:

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }\frac{5n^3-n}{n^2\left(n+2\right)\left(n^2+5\right)}}\le {\displaystyle \sum _{n-1}^{\infty }\frac{5}{n^2}}\\ <\infty \end{array}$

Therefore, the given series ${\displaystyle \sum _{n=1}^{\infty }\frac{5n^3-3n}{n^2\left(n+2\right)\left(n^2+5\right)}}$ converges.