Answer
$\dfrac{\pi}{2}$
Work Step by Step
Let us consider an infinite series $\Sigma_{n=k}^{n=\infty} a_n$. This series will converge to the sum $S$ when the sequence of its partial sums $\{S_n\}$ converge to $S$.
That is, $S=\lim\limits_{N \to \infty} S_n$ and $S=\Sigma_{n=k}^{n=\infty} a_n$.
This series will diverge when the limit does not exist and the series will diverge to infinity when the limit is infinite.
Here, we have
$S_n= \Sigma_{k=1}^{n} [\sin^{-1} (1/k) -\sin^{-1} (1/k+1)]$
or, $=[\sin^{-1} (1) -\sin^{-1} (1/2)]+[\sin^{-1} (1/2) -\sin^{-1} (1/3)]+.......+[\sin^{-1} (1/n) -\sin^{-1} (1/n+1)]$
or, $= \sin^{-1} (1) -\sin^{-1} (\dfrac{1}{n+1})$
Now, $S=\lim\limits_{n \to \infty} S_n=\lim\limits_{n \to \infty} [ \sin^{-1} (1) -\sin^{-1} (\dfrac{1}{n+1})]=\lim\limits_{n \to \infty} [\sin^{-1} (1)] -\lim\limits_{n \to \infty}\sin^{-1} (\dfrac{1}{n+1})$
or, $S=\dfrac{\pi}{2}-\sin^{-1} (\lim\limits_{n \to \infty}\dfrac{1}{n+1})$
or, $S=\dfrac{\pi}{2}-\sin^{-1}(0)=\dfrac{\pi}{2}$
