Convergence of difference of two divergent series

Question

Find the set of values of the real number $a$ for which $$\sum_{n=1}^{\infty} \left(\frac1{n} − \sin \frac1{n}\right)^a$$ converges.

The root and ratio test may not be of any help here. How do we approach this problem. Any hints? Thanks beforehand.

Answer 1

Hint. One has, as $ n \to \infty$, by a Taylor series expansion, $$ \left(\frac1{n} − \sin \frac1{n}\right)^a=\frac1{6^a}\frac1{n^{3a}}+O\left(\frac1{n^{3a+1}} \right) $$ then one may use a comparison test to conclude.