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?
 A: 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?
A: 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.
A: 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 $n$, $$n^2 > \frac{72C}{7}$$.
A: 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.
