# 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., $$\alpha>\frac{1}{3}$$.

• The limit comparison test is the way to go: if $a_n, b_n > 0$ and $a_n/b_n \to c \in (0, \infty)$ as $n\to \infty$ then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. Nov 26, 2020 at 4:52
• You can use the fact that the sin series is enveloping, so that $x > x-x^3/6+ x^5/120> \sin(x) > x-x^3/6$. Nov 26, 2020 at 6:17
• The power of $1/n$ in the majorant should be $3\alpha$ and not $\alpha/3$.
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!