For which $a>0$ series is convergent? 
For which $a>0$ series  $$\sum { \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n}  \right) \right)^a } $$ $(n \in \mathbb N)$ is convergent?

My try:From Taylor theorem I know that:$$a_{n}={ \left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n}  \right) \right)^a } = (\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a}$$
Then I have:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}}+o(\frac{1}{n^{8}}))^{a}$$At this point, my problem is that if I had:
$$(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}}))^{a} \le (\frac{1}{n^{4}})^{a}$$I could say that $0 \le a_{n} \le (\frac{1}{n^{4}})^{a}$ so for $a>\frac{1}{4}$ this series is convergent.However in this task I have also $o(\frac{1}{n^{8}}))^{a}$ and I don't know what I can do with it to finish my sollution.Can you help me?

The answer is indeed as you concluded $a > \frac{1}{4}$.
Note that (from the first line in your attempt)
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$
for $n$ sufficiently large, and that is all you need to conclude
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^{a} = \theta \left(\frac{1}{n^4} \right)^a = \frac{1}{n^{4a}},$$
as $\sum_{n=1}^{\infty} \theta \left(\frac{1}{n^4} \right)^a$ converges iff $a > \frac{1}{4}$. 
To elaborate:
$$\left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) = \theta\left(\frac{1}{n^4} \right),$$
which implies for some positive constants $C_1, C_2$
$$ \frac{C_1}{n^4} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right) \le \frac{C_2}{n^4} $$ 
which implies for positive $a$:
$$ \frac{C_1}{n^{4a}} \le \left(\frac{1}{n^{4}}-\frac{7}{72n^{6}}+o(\frac{1}{n^{8}})\right)^a \le \frac{C_2}{n^4a} $$ 
However, $\sum_{n=1}^{\infty} \frac{C_1}{n^{4a}}$ diverges for all positive $a \leq \frac{1}{4}$ so if $a$ is positive then $a$ must satisfy $a > \frac{1}{4}$. On the other hand $\frac{C_2}{n^{4a}}$ converges for all $a > \frac{1}{4}$ so it suffices that $a > \frac{1}{4}$.
Can use a similar line of reasoning to show that $a$ cannot be nonnegative for the sum to converge.
