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I am given the following fourier-series:

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\,\sin(nt)$$

I want to figure out if the series converges uniformly or pointwise on $\Bbb R$, and if converges pointwise for $t=\frac{\pi}{2}$:

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\,\sin\left(\frac{\pi}{2}n\right)$$

Can I use the Weierstrass-M-Test here?

$$\left\lvert\frac{(-1)^n}{n}\sin\left(\frac{\pi}{2}n\right)\right\rvert \le \left\lvert\frac{1}{n}\right\rvert$$

and since the the series $\sum_{n=0}^{\infty}\frac{1}{n}$ diverges my series is not uniformly convergent.

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  • $\begingroup$ Please take care to several points. (1) your definition of $f_n$ doesn't make sense (with your current definition the series is constant) and (2) you can't speak of uniform or pointwise convergence at a point. $\endgroup$ – mathcounterexamples.net Jun 7 '15 at 13:38
  • $\begingroup$ That's weird. It explicitly states on our problem sheet "investigate the behavior of $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\sin(nt)$ when $t=\frac{\pi}{2}$. Does the series converge uniformly or pointwise on $\Bbb R$". Are you saying this question doesn't make any sense? $\endgroup$ – qmd Jun 7 '15 at 13:43
  • $\begingroup$ Hint to explain (1) by Jean-Pierre: look for $n$s on both sides of an identity. $\endgroup$ – Did Jun 7 '15 at 13:45
  • $\begingroup$ @Did I am not sure what you mean. $\endgroup$ – qmd Jun 7 '15 at 13:49
  • $\begingroup$ Sorry but I do not "mean" anything, rather, I say that you should look for $n$ on both sides of the identity Jean-Pierre pointed you at. Do it. $\endgroup$ – Did Jun 7 '15 at 13:52
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Your series is the Fourier series of the function $f(x)=-\frac{x}{2}$ over $(-\pi,\pi)$, extended by periodicity. The proof is straighforward, you just have to compute $$ \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)\,dx $$ through integration by parts. Then the situation is the following:

$\hspace1in$ enter image description here

with pointwise convergence for every point of $I=(-\pi,\pi)$.

In virtue of Gibbs' phenomenon, the convergence on $I$ is not uniform.

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  • $\begingroup$ Thankyou very much. Is there another way to show convergence is not uniform on $I$? I am not familiar with the Gibb'S phenomenon. $\endgroup$ – qmd Jun 7 '15 at 14:07
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    $\begingroup$ @Suh: A simple way is to notice that $f(x)$ is not continuous over $\mathbb{R}$, while every partial sum of your series is. $\endgroup$ – Jack D'Aurizio Jun 7 '15 at 14:08
  • $\begingroup$ While to prove pointwise convergence you may use Carleson's theorem or just the fact that for any $x\in I$ your series is the imaginary part of $-\log(1+e^{ix})$, plus Dirichlet's test. $\endgroup$ – Jack D'Aurizio Jun 7 '15 at 14:13
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    $\begingroup$ Thanks again for your help. $\endgroup$ – qmd Jun 7 '15 at 14:16

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