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 term that is $o(n^{-8})$ is smaller, for all sufficiently large $n$, than $Cn^{-8}$, for at least some well-chosen value of $C$.  
Making use of that $C$, you can finish the proof by noting that for sufficiently large $n$, $\frac7{72n^6} > Cn^{-8}$ since for large enough $n$, $$n^2 > \frac{72C}{7}$$.
