TITLE: Determine for which α ∈ R the series is absolutely convergent.
QUESTION [5 upvotes]: Determine for which α ∈ R the series:
$$\sum_{n=1}^{\infty}\frac{\sin\big(e^{\sqrt{n+1}-\sqrt{n}} + \pi n -1\big)}{\ln(\arctan\frac{1}{n})}n^\alpha$$
is:
a) aboslutely convergent,
b) convergent,
c) divergent.
I've tried an approach where I considered the nominator and the denominator separately using direct comparison test for both and:

Taylor's theorem to approximate  $e^{\sqrt{n+1}-\sqrt{n}}$ as $e^x = x + 1 + o(x):$

$e^{\sqrt{n+1}-\sqrt{n}} = {\sqrt{n+1}-\sqrt{n}} + 1 + o(\sqrt{n+1}-\sqrt{n}) $

The fact that $\arctan\frac{1}{n} = \frac{π}{2} -\arctan{n} $

Neither of these helped me solve this problem. Any suggestions? Thanks for the help.

REPLY [2 votes]: Notice that
$$\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}\sim \frac{1}{2\sqrt{n}}$$
and since $\sin(x+n\pi) = (-1)^n\sin x$, we have an alternating series as well. That's it for the numerator, now looking closely at the denominator we have that
$$\log\left(\arctan\frac{1}{n}\right) \sim \log\left(\frac{1}{n}\right) = - \log n$$
meaning that (calling the above series $\sum a_n$) :
$$|a_n| \sim \frac{n^{\alpha - \frac{1}{2}}}{2\log n}$$
For $\alpha > \frac{1}{2}$ this will always diverge. At the very least for $\alpha < -\frac{1}{2}$ the series will converge absolutely. I will leave it to you to prove that statement, and figure out where else the series will absolutely or conditionally converge.