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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

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

1 Answer 1

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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 $3>1$.

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    $\begingroup$ Thank you for this answer. $\endgroup$
    – Srijan
    Sep 9, 2023 at 12:04

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