Finding the set of values of the real number $a$ for which $\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges 
Find the set of values of the real number $a$ for which $\displaystyle\sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a$ converges.

I had been stuck on this problem for a while but now I have managed to deduce following :
Using Maclaurin series for $\sin$ function,
$$\sin \frac 1n=\frac 1n - (\frac 1n)^3 \frac {1}{3!} + (\frac 1n)^5 \frac {1}{5!} - (\frac 1n)^7 \frac {1}{7!}+\dots$$
which implies
$$\frac 1n - \sin \frac 1n = (\frac 1n)^3 \frac {1}{3!} - (\frac 1n)^5 \frac {1}{5!} + (\frac 1n)^7 \frac {1}{7!}-\dots$$
and finally
$$0 \le (\frac 1n - \sin \frac 1n)^a \le ((\frac 1n)^3 \frac {1}{3!})^a$$
where the first inequality follows from the result : $\sin x \le x \; \forall x\gt 0$.
Hence
$$ 0 \le \sum_{n=1}^{\infty} (\frac 1n - \sin \frac 1n)^a \le \sum_{n=1}^{\infty} ((\frac 1n)^3 \frac {1}{3!})^a=\sum_{n=1}^{\infty} (\frac {1}{3!})^a \frac 1{n^{3a}}=(\frac {1}{3!})^a\sum_{n=1}^{\infty}\frac 1{n^{3a}}.$$
Last inequality converges for $3a \gt 1.$
Thus the given series converges for $a \in (\frac 13, \infty).$
Are there any loopholes in this solution? What are other ways to think about this problem?

For the divergence:
\begin{align*}
\sum_{n}\left(\frac{1}{n}-\sin\frac{1}{n}\right)^{a}&\geq\sum_{n}\left(\frac{1}{3!}\frac{1}{n^{3}}-\frac{1}{5!}\frac{1}{n^{5}}\right)^{a}\\
&=\frac{1}{(3!)^{a}}\sum_{n}\frac{1}{n^{3a}}\left(1-\frac{3!}{5!}\frac{1}{n^{2}}\right)^{a}\\
&\geq\frac{1}{(3!)^{a}}\frac{1}{2^{a}}\sum_{n\geq N}\frac{1}{n^{3a}}
\end{align*}
for large $N$, so it diverges for all $a\in(0,1/3]$. For $a\leq 0$ is clear, because the tail does not converge to zero.
