<s>[INST] 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?
 [/INST] It is much simpler to use equivalents, which can be found with Taylor's formula:


*

*$2(1-\cos x)=x^2-\dfrac{x^4}{12}+o(x^4)$,

*$\sin(\sin x)=\sin\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=\Bigl(x-\dfrac{x^3}6\Bigr)-\frac16\Bigl(x-\dfrac{x^3}6\Bigr)^{\!3}+o(x^3)=x-\dfrac{x^3}3+o(x^3)$, so
$$x\sin(\sin x)=x^2-\frac{x^4}3+o(x^4).$$
Now, replacing $x$ with $\frac 1n$, we obtain
$$2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left( \sin\frac{1}{n}  \right)=\frac1{n^2}-\frac1{12n^4}-\frac1{n^2}+\frac1{3n^4}+o\biggl(\frac1{n^4}\biggr)=\frac1{4n^4}+o\biggl(\frac1{n^4}\biggr)$$
so an asymptotic equivalent for the general term of the series is
$$\left(2-2 \cos\frac{1}{n} -\frac{1}{n}\cdot \sin\left(\sin\frac{1}{n}\right) \right)^a\sim_\infty\frac1{4^a n^{4a}}.$$
Knowing that series (with positive terms) which have asymptotically equivalent general terms both converge or both diverge, can you conclude?
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