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
Best Answer
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 $3>1$.