Chapter 8.4, Problem 8E

To determine

To check: The series is convergent or divergent.

Answer to Problem 8E

The comparison test the series ${\displaystyle \sum _{n=0}^{\infty }{(-1)}^n\frac{n}{\sqrt{n^3+2}}}$ is convergent series .

Explanation of Solution

Given information: ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\frac{n}{\sqrt{n^3+2}}}$

we know that ,

  $\begin{array}{l}{\displaystyle \sum _{n=k}^{\infty }{a}_n}\\ $ \text{ where }{a}_n={(-1)}^n\cdot b_n$\text{ and }{b}_n\ge 0\end{array}$

Then $\underset{n\to \infty }{\lim }{b}_n=0$ and $b_n$ decreasing sequence .

Then ${\displaystyle \sum _{n=k}^{\infty }{a}_n}$ is convergent here $\underset{n\to \infty }{\lim }{b}_n=\underset{n\to \infty }{\lim }\frac{n}{\sqrt{n^3+2}}=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{n+\frac{2}{n^2}}}$

Hence, by comparison test ${\displaystyle \sum _{n=1}^{\infty }{(-1)}^n\frac{n}{\sqrt{n^3+2}}}$ is convergent series .