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?


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ Change variable in the limit $x=1/n$. $\endgroup$ – Joe Feb 2 '15 at 17:04

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.