Question

If the sum $\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{2}{n^2+n+4}\right)$ is equal to ${\tan }^{-1}\left(\frac{a}{b}\right)$, where $a,b\in N$, then find the least value of $(a+b)$

Answer 1

Answer: 3

$\sum _{n=1}^{\infty }{\tan }^{-1}\left(\frac{2}{n^2+n+4}\right)=\sum _{n=1}^{\infty }{\tan }^{-1}\left\{\frac{\frac{2}{n(n+1)}}{1+\frac{4}{n(n+1)}}\right\}$
$=\sum _{n=1}^{\infty }{\tan }^{-1}\left\{\frac{\frac{2}{n}-\frac{2}{n+1}}{1+\frac{2}{n}\cdot \frac{2}{n+1}}\right\}=\sum _{n=1}^{\infty }\left\{{\tan }^{-1}\left(\frac{2}{n}\right)-{\tan }^{-1}\left(\frac{2}{n+1}\right)\right\}$
$=\left\{{\tan }^{-1}\left(\frac{2}{1}\right)-{\tan }^{-1}\left(\frac{2}{2}\right)\right\}+\left\{{\tan }^{-1}\left(\frac{2}{2}\right)-{\tan }^{-1}\left(\frac{2}{3}\right)\right\}+\dots \dots \infty \text{ terms }$
$={\tan }^{-1}2={\tan }^{-1}\left(\frac{a}{b}\right)$
$\Rightarrow a=2;b=1\Rightarrow (a+b)=3$