Answer
$-\dfrac{\pi}{4}$
Work Step by Step
The nth partial sums are:$ s_n=(\tan ^{-1} (1) - \tan ^{-1} (2))+(\tan ^{-1} (2) - \tan ^{-1} (3))+.......+(\tan ^{-1} (n) - \tan ^{-1} (n+1))=\tan ^{-1} (1) - \tan ^{-1} (n+1)$
Now, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [(\tan ^{-1} (1) - \tan ^{-1} (n+1))]=\tan^{-1} (1)-(\dfrac{\pi}{2})$
or, $=\dfrac{\pi}{4}-\dfrac{\pi}{2}$
or, $=-\dfrac{\pi}{4}$
