# Convergence of series $\sum\limits_{n=1}^\infty\sin\frac{1}{n^{3}}$

Problem:

I am trying to prove that series $$\sum\limits_{n=1}^\infty\sin\frac{1}{n^{3}}$$ converges.

My attempt:

$$\sum\limits_{n=1}^\infty|\sin\frac{1}{n^{3}}| \leq \sum\limits_{n=1}^\infty\frac{1}{n^3}$$. Then $$\sum b_n = \sum 1/n^3$$ is convergent $$p$$ series, where $$p = 3$$. Hence by comparison test given series in the question converges.

or

If $$a_n = \sin \frac{1}{n^3}$$ and $$b_n = 1/n^3%$$ then $$\lim \frac{a_n}{b_n} = 1$$, as $$n \rightarrow\infty$$. Hence by the limit comparison test both the series converges.

Am I on the right path to solve this question?, or are there alternative approaches to address this problem? Thank you for your assistance.

Thanking you

• Both of these are good options for answering the question, and both solutions look good to me. Sep 9, 2023 at 11:45
• Did you forget the sum-sign on the right side ? Apart from this , the solution is correct. Sep 9, 2023 at 11:49
• @Peter I forgot that. Thank you for bringing it to my attention. Sep 9, 2023 at 11:52
• @MatthewLeingang Thanks for the comment. Sep 9, 2023 at 11:52

Your solutions are okay, but you can also use the Taylor approximation of the sine function: $$\sin x\sim x$$ when $$x\to 0$$. $$\sum\limits_{n=1}^\infty\sin\frac{1}{n^{3}}\sim \sum\limits_{n=1}^\infty \frac{1}{n^3}$$ and the last series converges since $$3>1$$.

• Thank you for this answer. Sep 9, 2023 at 12:04