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

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!
