Chapter 10.5, Problem 15E

To determine

To find: that the whether the series converges or diverges.

Answer to Problem 15E

Converges

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+1}}$

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+1}<}{\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2}}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2}}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^{3/2}}}\end{array}$

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^{3/2}}}$ is convergent because any series of the from ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ where p > 1 is convergent

Therefore, ${\displaystyle \sum _{n=1}^{\infty }\frac{\sqrt{n}}{n^2+1}}$ is convergent