Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I encountered this problem while studying for an analysis exam. Here is a related question I asked some days ago.
The problem is as follows: Suppose $a_n$ is a decreasing sequence of positive real numbers and that$$\sum_{n = 0}^{\infty}{a_n \sin{(nx)}}$$ converges uniformly on $\mathbb{R}$, show that $$\lim_{n \to \infty}{(n a_n)} = 0.$$ Any tip or solution is welcome, and also avoid using Fourier series, because they haven't been introduced in the book so it can be solved without using them.

share|cite|improve this question
This question does not only seem to be related to the one you mention, but actually identical! – wildildildlife Mar 26 '11 at 20:25
@wildildildlife: No, in the previous question I asked wether $\sum{\frac{\sin(nx)}{n}}$ converge uniformly on the real numbers. The question was inspired by this problem but I never asked the solution to the problem. There was some discussion in the comments about the possible solution, but I couldn't solve it so that is why I'm asking. – alejopelaez Mar 26 '11 at 20:32
@arturo: Like I told wildildildlife, the questions are different. In the other one I asked about the convergence of $\sum{\frac{\sin{(nx)}}{n}}$ in order to get some insight on this problem. But seeing I wasn't able to solve it, I raised this question. They are related because this problem inspired my previous question, but they are not the same. – alejopelaez Mar 26 '11 at 20:38
@Arturo, @wildildildlife: Seems to me Peláez is right -- the (accepted) answer to the other question doesn't answer this question, and the discussion in the comments there also doesn't resolve it. Although this question was used almost verbatim to motivate the other question, it was neither asked nor answered there. There's a strange asymmetry in the software -- I can vote to close this question, but it seems I can't vote against closing it? – joriki Mar 26 '11 at 20:42
@Peláez: Fair enough. Do please note that one does not get messages updating comments like you do for questions (a banner saying "this has been edited" and "a question has been posted"). Even though you made your reply to wildildildlife before I voted to close on that basis, I had not seen your reply (had not seen your comment until now when I reloaded). So chiding me with "Like I told..." is not really appropriate. – Arturo Magidin Mar 26 '11 at 20:43
up vote 10 down vote accepted

$\sum_{i=[(k+1)/2]}^k a_i \sin(ix)$ goes uniformly to 0 as $k\to\infty$. Set $x=\pi/(2k)$. Then all $a_i$'s are $\geq a_k$, all the sines are $\geq1/\sqrt{2}$, hence the sum is $\geq (k-1)/2\times a_k/\sqrt{2}$. Since this goes to 0, so does $k a_k$

share|cite|improve this answer
+1: nice succinct answer :-) – joriki Mar 26 '11 at 21:24
@user8268 could you explain your answer please? Explain the motivation for letting $i = \frac{k+1}{2}$, how $a_i \geq a_k$, etc. Thanks – Ozera Feb 25 '15 at 23:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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