# Explore the convergence of the series: sum_{n=1}^infty ncdotarctan^2left(frac{pi}{8n+1}right)

Explore the convergence of the series:
$\sum _{n=1}^{\mathrm{\infty }}n\cdot {\mathrm{arctan}}^{2}\left(\frac{\pi }{8n+1}\right)$
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Aniqa O'Neill

Given, the series $\sum _{n=1}^{\mathrm{\infty }}n\cdot \left[{\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right]}^{2}$

We will use limit comparison test to test the convergence of it.

Limit Comparison Test:

For the two series $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ and $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ with ${a}_{n}\ge 0,{b}_{n}>0,\mathrm{\forall }n$.

Define $c=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}$.

If 0, then both the series converge or diverge together.

In our case, ${a}_{n}=n\left[{\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right]}^{2}$

We know that, ${\mathrm{tan}}^{-1}\left(\frac{1}{x}\right)\sim \frac{1}{x}$ for large values of x.

So, we have

${\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right)\sim \frac{1}{n}$

$⇒\left({\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right)}^{2}\sim \frac{1}{{n}^{2}}$

$⇒n\left({\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right)}^{2}\sim \frac{n}{{n}^{2}}=\frac{1}{n}$

This suggests we consider ${b}_{n}=\frac{1}{n}$ whose series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$ diverges.

By Limit comparison test,

$\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}$

$=\underset{n\to \mathrm{\infty }}{lim}\frac{n\left[{\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right]}^{2}}{\frac{1}{n}}$

$=\underset{n\to \mathrm{\infty }}{lim}{n}^{2}\left[{\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right]}^{2}$

We used WolframAlpha calculator to obtain its value,

$=\frac{{\pi }^{2}}{64}$

Hence we have $c=\frac{{\pi }^{2}}{64}$ which lies between 0 and $\mathrm{\infty }$

Therefore, by the limit comparison test, as $\sum _{n=1}^{\mathrm{\infty }}{b}_{n}$ diverges, the series

$=\sum _{n=1}^{\mathrm{\infty }}n\cdot \left[{\mathrm{tan}}^{-1}\left(\frac{\pi }{8n+1}\right){\right]}^{2}$ must diverge.

Jeffrey Jordon