Convergence of $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$

Question

I am trying to seek the real values of $\alpha$ for convergence of the following series : $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha}$

My trial: Since $n>1$, $0<\frac{1}{n}<1$, thus we can expand sin function as \begin{align} \sin\left(\frac{1}{n} \right) = \frac{1}{n} - \frac{1}{3!} \frac{1}{n^3} + \cdots \end{align} Hence the series becomes \begin{align} \sum_{n=1}^{\infty} \left(\frac{1}{n} - \sin\left(\frac{1}{n} \right) \right)^{\alpha} = \sum_{n=1}^{\infty} \left(\frac{1}{3!}\frac{1}{n^3} - \cdots\right)^{\alpha} \leq \frac{1}{3!}\sum_{n=1}^{\infty} \left(\frac{1}{n}\right)^{3\alpha} \end{align} Thus my guess is from p-test $3\alpha >1$, i.e., $\alpha>\frac{1}{3}$.

Is this approach admissible?

Answer 1

Your intuition is correct for using the Taylor expansion of $\sin x$ around $x=0$. Note that for large enough $n$: $$ {1\over n}-{1\over 6n^3}\le\sin {1\over n}\le {1\over n}-{1\over 5n^3} $$ You have already done the lower bound. The upper bound is just on its way!