Chapter 8.3, Problem 16E

To determine

To find: whether the series is convergent or divergent.

Answer to Problem 16E

The series is convergent.

Explanation of Solution

Given information: the series is ${\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{n^3+1}}$

Theorem: The necessary condition for convergence of the series ${\displaystyle \sum u_n}$ is that $u_n\to 0\text{ as }n\to \infty$ .

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

Let, $u_n={\displaystyle \sum _{n=1}^{\infty }\frac{n^2}{n^3+1}}$

Also let $a_n=\frac{n^2}{n^3+1}$

Now we check the limit of the sequence of the partial sum take limit both side, we have

  $\begin{array}{l}\Rightarrow \underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{n^2}{n^3+1}\\ \Rightarrow \underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{\frac{1}{n}}{1+\frac{1}{n}}\\ \Rightarrow \underset{x\to \infty }{\lim }{a}_n=0\end{array}$

From the above theorem we say that the series is convergent.