Question

If the sum $\displaystyle\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{2}{n^2+n+4}\right)$ 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

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