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?

  • $\begingroup$ I think your approach works but you have to prove it rigorously . $\endgroup$
    – DeepSea
    Nov 26, 2020 at 4:22
  • $\begingroup$ @DeepSea, that's the problem! I want to know what process is needed for rigorous proof! $\endgroup$
    – phy_math
    Nov 26, 2020 at 4:33
  • 2
    $\begingroup$ 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. $\endgroup$
    – user307205
    Nov 26, 2020 at 4:52
  • $\begingroup$ You can use the fact that the sin series is enveloping, so that $x > x-x^3/6+ x^5/120> \sin(x) > x-x^3/6$. $\endgroup$ Nov 26, 2020 at 6:17
  • $\begingroup$ The power of $1/n$ in the majorant should be $3\alpha$ and not $\alpha/3$. $\endgroup$
    – Gary
    Nov 26, 2020 at 7:05

1 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!


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