Question

The sum $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{2}{n^2}\right)$ equals $\frac{\pi }{m}$.Find $m$

Answer 1

Given $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2n^2}\right)=\frac{\pi }{m}$

Let us consider ${\tan }^{-1}\left(\frac{1}{2n^2}\right)={\tan }^{-1}\left(\frac{1\times 2}{2n^2\times 2}\right)$

$={\tan }^{-1}\left[\frac{2}{4n^2}\right]$

$={\tan }^{-1}\left[\frac{2}{1+4n^2-1}\right]$

$={\tan }^{-1}\left[\frac{2}{1+(2n+1)(2n-1)}\right]$

$={\tan }^{-1}\left[\frac{(2n+1)-(2n-1)}{1+(2n+1)(2n-1)}\right]$

$={\tan }^{-1}(2n+1)-{\tan }^{-1}(2n-1)\left[{\tan }^{-1}\right[\frac{x-y}{1+xy}={\tan }^{-1}x-{\tan }^{-1}y]$

$\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2n^2}\right)=\sum _{n=1}^{\infty }{\tan }^{-1}(2n+1)-\sum _{n=1}^{\infty }{\tan }^{-1}(2n-1)$

$=\left[{\tan }^{-1}\right(3)-{\tan }^{-1}(1\left)\right]+\left[{\tan }^{-1}\right(5)-{\tan }^{-1}(3\left)\right]+\left[{\tan }^{-1}\right(7)-{\tan }^{-1}(5\left)\right]+\left[{\tan }^{-1}\right(2n+1)-{\tan }^{-1}(2n-1\left)\right]$

$={\tan }^{-1}(2n+1)-{\tan }^{-1}\left(1\right)$

$=\underset{n\to \infty }{Lt}\left[{\tan }^1\right(2n+1\left){\tan }^{-1}\right(1\left)\right]$

$={\tan }^{-1}\left(\infty \right)-{\tan }^{-1}\left(1\right)$

$=\frac{\pi }{2}-\frac{\pi }{4}$

$=\frac{\pi }{4}$

Given $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{1}{2n^2}\right)=\frac{\pi }{m}=\frac{\pi }{4}$