Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.sum_{n=1}^inftyfrac{arctan n}{n^2+1}

To Determine:
Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
Given: we have $\sum _{n=1}^{\mathrm{\infty }}\frac{{\tan }^{-1}n}{n^2+1}$
Explanation: integral test applies when the function is positive and increasing for $n\ge N$
we can write it as
${\int }_1^{\mathrm{\infty }}\frac{{\tan }^{-1}x}{x^2+1}dx=\underset{a\to \mathrm{\infty }}{lim}{\int }_1^a\frac{{\tan }^{-1}x}{x^2+1}$
let,
$u={\tan }^{-1}x$
$du=\frac{1}{x^2+1}$
when x=1 then $u={\tan }^{-1}x\Rightarrow u=\frac{\pi }{4}$
when x=a then $u={\tan }^{-1}a$
the integral becomes
$\underset{a\to \mathrm{\infty }}{lim}{\int }_{\frac{\pi }{4}}^{{\tan }^{-1}a}udu=\underset{a\to \mathrm{\infty }}{lim}{\left[\frac{u^2}{2}\right]}_{\frac{\pi }{4}}^{{\tan }^{-1}a}$
$=\underset{a\to \mathrm{\infty }}{lim}\frac{1}{2}({\tan }^{-1}a{)}^2-\frac{1}{2}(\frac{\pi }{4}{)}^2$
$[\underset{a\to \mathrm{\infty }}{lim}{\tan }^{-1}a={\tan }^{-1}\mathrm{\infty }=\frac{\pi }{2}]$
$\underset{a\to \mathrm{\infty }}{lim}{\int }_{\frac{\pi }{4}}^{{\tan }^{-1}a}udu=\frac{1}{2}((\frac{\pi }{2}{)}^2-((\frac{\pi }{4}{)}^2))$
$=\frac{1}{2}(\frac{{\pi }^2}{4}-\frac{{\pi }^2}{16})$
$=\frac{3{\pi }^2}{32}$
here we see that the series converges.