Chapter 8.3, Problem 18E

To determine

To find: whether the series is convergent or divergent.

Answer to Problem 18E

The series is convergent.

Explanation of Solution

Given information: the series is ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2+9}}$

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 ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2+9}}$

Let, $u_n={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^2+9}}$

Also let $a_n=\frac{1}{n^2+9}$

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

  $\begin{array}{l}\Rightarrow \underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }\frac{1}{n^2+9}\\ \Rightarrow \underset{x\to \infty }{\lim }{a}_n=\underset{x\to \infty }{\lim }{\left(\frac{1}{n}\right)}^2·\frac{1}{(1+{\left(\frac{3}{n}\right)}^2)}\\ \Rightarrow \underset{x\to \infty }{\lim }{a}_n=0\end{array}$

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