 beljuA

2020-11-08

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.
$\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{arctan}n}{{n}^{2}+1}$ hesgidiauE

Gived:
Verify if the series may be used with the Integral Test. The Integral Test can then be used to assess if the series is converging or diverging.
Given: we have $\sum _{n=1}^{\mathrm{\infty }}\frac{{\mathrm{tan}}^{-1}n}{{n}^{2}+1}$
The integral test is applicable when the function is positive and rising for the reason that $n\ge N$
we can write it as
${\int }_{1}^{\mathrm{\infty }}\frac{{\mathrm{tan}}^{-1}x}{{x}^{2}+1}dx=\underset{a\to \mathrm{\infty }}{lim}{\int }_{1}^{a}\frac{{\mathrm{tan}}^{-1}x}{{x}^{2}+1}$
let,
$u={\mathrm{tan}}^{-1}x$
$du=\frac{1}{{x}^{2}+1}$
when x=1 then $u={\mathrm{tan}}^{-1}x⇒u=\frac{\pi }{4}$
when x=a then $u={\mathrm{tan}}^{-1}a$
the integral becomes
$\underset{a\to \mathrm{\infty }}{lim}{\int }_{\frac{\pi }{4}}^{{\mathrm{tan}}^{-1}a}udu=\underset{a\to \mathrm{\infty }}{lim}{\left[\frac{{u}^{2}}{2}\right]}_{\frac{\pi }{4}}^{{\mathrm{tan}}^{-1}a}$
$=\underset{a\to \mathrm{\infty }}{lim}\frac{1}{2}\left({\mathrm{tan}}^{-1}a{\right)}^{2}-\frac{1}{2}\left(\frac{\pi }{4}{\right)}^{2}$
$\left[\underset{a\to \mathrm{\infty }}{lim}{\mathrm{tan}}^{-1}a={\mathrm{tan}}^{-1}\mathrm{\infty }=\frac{\pi }{2}\right]$
$\underset{a\to \mathrm{\infty }}{lim}{\int }_{\frac{\pi }{4}}^{{\mathrm{tan}}^{-1}a}udu=\frac{1}{2}\left(\left(\frac{\pi }{2}{\right)}^{2}-\left(\left(\frac{\pi }{4}{\right)}^{2}\right)\right)$
$=\frac{1}{2}\left(\frac{{\pi }^{2}}{4}-\frac{{\pi }^{2}}{16}\right)$
$=\frac{3{\pi }^{2}}{32}$
In this instance, the series converges. Jeffrey Jordon