Arcsin Series : $\sum\limits_{n=1}^{\infty}\Bigl [\frac {\pi}{2} -\arcsin\bigl(\frac{n}{n+1}\bigr)\Bigr]^{\alpha}$

Question

Good morning everyone,

I'd like to discuss with you the following exercise :

$$\sum\limits_{n=1}^{\infty}\Bigl[\frac {\pi}{2} - \arcsin\Bigl(\frac{n}{n+1}\Bigr)\Bigr]^{\alpha}$$

After verified the divergence test, which gave $\alpha > 0 $ I tried to evaluate the series or at least shrink it between its "highest" and "lowest" values.

The result didn't make me happier because I found :

$0 < f(x) < \pi$, where $f(x)= \bigl[\frac {\pi}{2} - \arcsin(\frac{n}{n+1})\bigr]^{\alpha}$.

I feel the result somehow incorrect, because I can't find any sort of method to make it converge or diverge.

Could anyone help me ?

Would be appreciated,thanks anyway.

Answer 1

$$\arcsin\frac{n}{n+1} = \arctan \frac{n}{\sqrt{2n+1}}=\frac{\pi}{2}-\arctan\frac{\sqrt{2n+1}}{n}$$ so the given series is convergent for any $\alpha>2$ and divergent for any $\alpha\leq 2$ by asymptotic comparison ($\arctan\frac{\sqrt{2n+1}}{n}\sim \sqrt{\frac{2}{n}}$) and the p-test.