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So my teacher gave a pretty tough (for me) problem in class today: Does the series $$ \sum_{n=1}^{\infty}\frac {\sin{\frac 1 n}} {\sqrt n} $$ converge or diverge?

So far I've thought of trying the Comparison Test and Ratio Test but couldn't really get anywhere. Intuitively I think it should converge, but how do I actually show this?

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Since $\sin(1/n)\sim_{+\infty}1/n$ then the series converges iff $\frac{1/n}{\sqrt n}=\frac{1}{n^{3/2}}$ is the general term of a convergent series. Being $3/2>1$, that's the case, so your series converges.

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  • $\begingroup$ Thanks, this was a nice solution. But how did you know $sin(1/n)$ approaches $1/n$? Is it part of a more general rule, or is it more of a fact one should know in and of itself? $\endgroup$ – Asker Feb 2 '15 at 16:30
  • $\begingroup$ It's simply the fundamental limit $\lim_{x\to0}\frac{\sin x}{x}=1$. $\endgroup$ – Joe Feb 2 '15 at 16:46
  • $\begingroup$ Can you explain how one gets there from that? (I don't doubt it's right, I just still don't understand) $\endgroup$ – Asker Feb 2 '15 at 16:52
  • $\begingroup$ It's the asymptotical criterion for series convergence. Suppose you have to check the convergence of $\sum_{n\ge0}a_n$, but you can't handle the sequence $(a_n)_n$ successfully (e.g. ratio and root criteria fail). If you find another sequence $(b_n)_n$ such that $\lim_na_n/b_n=\lambda\in\Bbb R,\;\;\lambda\neq0$ then the two series $\sum_{n\ge0}a_n$ and $\sum_{n\ge0}b_n$ will have the same convergence character. $\endgroup$ – Joe Feb 2 '15 at 16:58
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    $\begingroup$ Change variable in the limit $x=1/n$. $\endgroup$ – Joe Feb 2 '15 at 17:04
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It suffices to use the fact that $|\sin x|\le|x|$. (See here or here or or here. You can probably find this inequality in many other places.)

Then you get $|a_n| \le \frac1{n\sqrt n}$ for your series, which means that it is absolutely convergent.

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