# Does $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$ in $[0,2\pi]$

This question arises because of a problem I was doing (Bartle 3rd edition, section 9.4 problem 3). It was like this.
Given $a_n$ a decreasing sequence of positive numbers and suppose that $$\sum_{n=0}^{\infty}{a_n \sin{(nx)}}$$ Converge uniformly (It doesn't specify the domain, so I guess is for every x). Prove that $n a_n \to 0$.
Clearly $\frac{1}{n}$ fits the description of $a_n$, and $n \frac{1}{n} \to 1 \neq 0$, so this would prove that there is a mistake in the problem if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for all $x$.
So my question is if $\sum{\frac{\sin{(nx)}}{n}}$ converge uniformly for every $x$.

(I know that the series converge uniformly for every x in $[\delta, 2\pi - \delta]$, for $0 < \delta <2\pi$ by using the Dirichlet criterion.)

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Use $\backslash$sin for the sine function. –  Did Mar 24 '11 at 18:26
You don't mean to say that the $a_n$ are integers. –  Douglas Zare Mar 24 '11 at 19:31
Please let us know if you come up with something for the original exercise. I'm curious! Have you tried something? –  Giuseppe Negro Mar 24 '11 at 19:49
@Douglas: Thanks, I changed that, $a_n$ can be any decreasing sequence of positive reals. –  alejopelaez Mar 24 '11 at 22:36
@dissonance: I was thinking of proving that $\sum{n a_n}$ converge, and that would imply $n a_n \to 0$. For that I was thinking of using the Cauchy criterion to bound the partial sums of $\sum{n a_n}$. But I'm thinking that $\sum{n a_n}$ doesn't necessarily converge. Another approach was using the fact that $a_n \sin{(nx)} \to 0$ uniformly, so I tried to pick an x that would allow me to relate it to $n a_n$. But so far I haven't been able to solve it. –  alejopelaez Mar 24 '11 at 22:54

The sum of the series is non-continuous (you can view this as the Fourier series for a saw-tooth function; or just check the behavior around x=0), so the convergence cannot be uniform. Each summand is obviously a continuous function, and a uniformly convergent series of continuous functions is continuous.

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What behavior around x=0 are you referring to? –  Digital Gal Mar 27 '11 at 4:24
The sum is positive and bounded away from 0 for small positive x, and negative for small negative x. I think this can be shown "by hand" without resorting to Fourier analysis, in case the OP is not familiar with it. –  Alon Amit Mar 27 '11 at 5:05
Actually, what we want to look at is near $2\pi$, not $0$, right? –  Pedro Tamaroff May 22 '13 at 18:07
What's the difference? –  Alon Amit May 23 '13 at 19:04

Hint:

Use Cauchy Criterion to prove that your infinite series isn't uniformly converges for all $x\in[0,2\pi]$.

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