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I am studying the convergence of the series: $$\sum\limits_{n=1}^\infty n^{3/2}\sin\left(\frac{3}{n^3}\right) $$

I have tried the following: As, $|\sin x|\le 1$ and term of the series are positves we have, $$\sum\limits_{n=1}^\infty n^{3/2}\sin\left(\frac{3}{n^3}\right) \le \sum\limits_{n=1}^\infty n^{3/2} =\infty$$

but this estimate leads neither to convergence nor to the divergence.

What is an elegant to prove the convergence or the divergence of this series? thanks in advance.

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    $\begingroup$ use that $|\sin(x)|\leq |x|$. $\endgroup$ – Kelenner Nov 21 '17 at 16:51
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As $|\sin(x)|\le |x|$ and as in your series you are feeding the $\sin$ positive numbers you get

$$\sum_n n^{3/2}\sin\left(\frac{3}{n^3} \right)\le 3\sum_n \frac{1}{n^{3/2}}\,, $$

which is known to converge.

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$$\frac{n^{3/2}\sin\frac3{n^3}}{\frac{n^{3/2}}{n^3}}=3\,\frac{\sin\frac3{n^3}}{\frac3{n^3}}\xrightarrow[n\to\infty]{}3$$

and thus, by the limit comparison test, our series converges since $\;\sum\limits_{n=1}^\infty\frac{n^{3/2}}{n^3}=\sum\limits_{n=1}^\infty\frac1{n^{3/2}}\;$ converges.

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  • $\begingroup$ Can you please take a look to MY answer and kindly explain to me what is wrong, because it looks exactly like your answer. You would do me a great favor because another user is tormenting me since a long time saying that my answers are wrong, even if they are right. Thank you in advance $\endgroup$ – Raffaele Nov 22 '17 at 10:19
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    $\begingroup$ @Raffaele It honestly does not look as your answer, and even less "exactly" . Please read my comment under your answer. And sometimes some users here can be tough, specially when trying to correct somebody else's answers, but in this case Did's critic is correct. $\endgroup$ – DonAntonio Nov 22 '17 at 11:33

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