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?

Use that you can find constants $0<c, C$ such that $c/n^4 \leq 1/n^4 -7/(72 n^6)+ o(1/n^8) \leq C /n^4$. Then you use minorant and majorant test to conclude that the series over the middle one converges iff the series over $1/n^4$ converges.
