Find the sum of arcsin 1/√2 + arcsin (√2 - 1)/√6 + ... + arcsin {√n - √(n - 1)}/√{n(n + 1)}, for n approaching infinity?
I suggest first checking that the series converges.
Calcuating a_n for a few large n suggests that the ratio test would be inconclusive (ratio = 1).
Using Raabe's test for few large n suggests that the series may be divergent. However, while Raabe's test can be used to show that a series is convergent I am not sure that the converse is true.
I suspect that one of the convergence tests in your toolbox may show that this series diverges.
let $\displaystyle a_n=\arcsin \left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right).$ then: $\displaystyle \sin a_n=\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n(n+1)}}$ and hence: $\displaystyle \cos a_n=\frac{1+\sqrt{n(n-1)}}{\sqrt{n(n+1)}}.$ therefore: $\displaystyle \tan a_n=\frac{\sqrt{n}-\sqrt{n-1}}{1+\sqrt{n(n-1)}}.$ thus:
$\displaystyle a_n=\arctan \left(\frac{\sqrt{n} - \sqrt{n-1}}{1+\sqrt{n(n-1)}}\right)=\arctan \sqrt{n} - \arctan \sqrt{n-1}.$ hence: $\displaystyle \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}(\arctan \sqrt{n} - \arctan \sqrt{n-1})=\lim_{n\to\infty} \arctan \sqrt{n}=\frac{\pi}{2}. \ \ \ \ \ \square$