Q:

Find all $\alpha$ such that the series converges

Find all values of $\alpha$ such that series $$\sum^\infty_{n=1} \left( \frac{1}{n \cdot \sin(1/n)} - \cos\left(\frac{1}{n}\right) \right)^\alpha$$ converges. 

I used Maclaurin for $\sin$ and $\cos$ and got:
$$a_n = \left( \frac{1}{1 - \dfrac{1}{3!n^2} + \ldots} - 1 + \frac{1}{2!n^2} - \frac{1}{4!n^4} + \ldots \right) ^ \alpha$$
Put it together in one fraction seems to be a hard thing to do. 

A:

HINT:
You are on the right track.   Note that 
$$\frac{1}{1-\frac1{6n^2}+O\left(\frac1{n^4}\right)}=1+\frac1{6n^2}+O\left(\frac1{n^4}\right)$$
Hence, we see that
$$\frac{1}{n\sin(1/n)}-\cos(1/n)=\frac2{3n^2}+O\left(\frac1{n^4}\right)$$

