Infinite Tangent....

It is an old IIT-JEE problem. A beautiful example of telescoping series!!

On observation one can easily write the general term of the sequence given in the question. $S = \sum_{i=1}^{\infty} \tan^{-1}\dfrac{1}{2i^{2}}$ Now we will try to convert the general term into telescoping sum. $S = \sum_{i=1}^{\infty} \tan^{-1}\dfrac{2}{1+4i^{2}-1}$ $S = \sum_{i=1}^{\infty} \tan ^{-1}\dfrac{2i+1-(2i-1)}{1+(2i+1)(2i-1)}$ Now we will apply the formula from inverse trigonometry that:- $\tan ^{-1} \Bigg(\dfrac{A-B}{1+AB}\Bigg) = \tan^{-1}A-\tan^{-1}B$ Hence our Sum becomes:- $S= \sum_{i=1}^{\infty} \tan^{-1}(2i+1) - \tan^{-1}(2i-1)$ We can easily observe that the sum is now telescoping. $S = \lim_{n\to\infty} (tan^{-1} 3 -\tan^{-1} 1+\tan^{-1} 5-\tan^{-1} 3 \ldots \tan^{-1} (2n+1) - \tan^{-1}(2n-1)$ After cancelling the corresponding terms we get:- $S = \lim_{n\to\infty} \tan^{-1} (2n+1) - \tan^{-1} 1$ $S = \dfrac{\pi}{2} - \dfrac{\pi}{4}$ $S = \dfrac{\pi}{4}$ Hence the answer is $\boxed{4}$ .