Answer
Converges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{1 }{n^2+1}$$
Since $f(n)= \frac{1 }{n^2+1},\ \ f'(n) =-\frac{2n}{\left(n^2+1\right)^2}$, is positive, decreasing and continuous for $n\geq1$, then by using the integral test
\begin{aligned} \int_{1}^{\infty} f(n) dn &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1 }{n^2+1}d n \\
&=\left.\lim _{t \rightarrow \infty}\tan^{-1}n\right|_{1} ^{t} \\
&= \lim _{t \rightarrow \infty} (\tan^{-1}t-\tan^{-1}1)\\ &=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}\end{aligned}
Thus the series converges.
