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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.

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alejopelaez
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  • 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
  • @Joriki: Yes: I only read the first paragraph of the previous question, which was identical to this, and got misled. Alas, there is no "unvote" for the close thing. – Arturo Magidin Mar 26 '11 at 20:44
  • @Arturo: Sorry, didn't mean to sound disrespectful, I know that the two questions may seem similar and that is why I answer your comment after having answered to wildildildlife. Also English is not my mother language so a lot of expressions that I use may seem rude, but is just because lack of words. Again sorry. – alejopelaez Mar 26 '11 at 21:05
  • @Peláez: No harm done. (English is not my native language either...) – Arturo Magidin Mar 26 '11 at 21:27

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$\sum_{i=[(k+1)/2]}^k a_i \sin(ix)$ 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 $k a_k$

user8268
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  • @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