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.

Hint
$$n\sin(\frac{1}{n})=1-\frac{1}{6n^2}(1+\epsilon_1(n))$$
$$\frac{1}{n\sin(\frac{1}{n})}=1+\frac{1}{6n^2}(1+\epsilon_2(n))$$
$$\cos(\frac{1}{n})=1-\frac{1}{2n^2}(1+\epsilon_3(n))$$
thus, when $n\to +\infty$, the general term of your series $u_n$, satisfies
$$u_n \sim (\frac{2}{3n^2})^\alpha $$
and by the limit comparison test, 
$\sum u_n$  converges $\iff \;
\alpha>\frac{1}{2}$.
