Solutions - 0180-01

$$a_n=\frac{1}{n^2+1}\gg \gg a_x=\frac{1}{x^2+1}$$

First note that:

1. $a_x$ is continuous
2. $a_x$ is positive
3. $a_x$ is monotone decreasing because $x^2+1$ is increasing

Then:

$$\begin{array}{c}{\int }_1^{\infty }\frac{1}{x^2+1}dx\gg \gg \underset{R\to \infty }{\lim }{\int }_1^R\frac{1}{x^2+1}dx\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\tan }^{-1}(x){|}_1^R\\ \\ \gg \gg \underset{R\to \infty }{\lim }{\tan }^{-1}(R)-\frac{\pi }{4}\\ \\ \gg \gg \frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\end{array}$$

Since this is finite, the integral test establishes that the series converges.