2021/107) Show that $\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right )$ - and deduce that $\sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}$

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$

Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$

Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$

We get  $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)-  \tan ^{-1}n$

Adding from 0 to k-1 we get as telescopic sum

Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} =  \tan ^{-1}k-  \tan ^{-1}0 = \tan ^{-1}k$

Taking limit as $k = \infty$

$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1}  = \tan ^{-1}\infty= \frac{\pi}{2}$