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For which $p>0$ is the series $L^2$-convergent?

$$\sum_{n=1}^ ∞ \frac{1}{n^p}e^ {ix} $$

Now I've been trying to solve this but just cant get the right answer. Another dude on university told me that the answer should be $p>1/2$ and he was pretty sure of it. He told me they couldnt solve it until they got a Theorem that doesnt exist in the "Fourier Analysis.."-book written by Gerald B.Folland.

I did a couple of tries like this but didnt come to anything good. Anyone who think differently and can help me solve this?

Jean-Claude Arbaut
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kamikazi
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  • Which $L^2$ space? $L^2([0,2\pi])$? $L^2([0,1])$? … – Daniel Fischer Sep 28 '16 at 18:37
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    Might want to double check that sum - do you mean $\sum_{n\ge1} \frac{1}{n^p} e^{inx}$? – πr8 Sep 28 '16 at 18:38
  • @DanielFischer Well, It doesnt say, but in the previous question (1a), it wanted us to prove that it's $L^2(-pi,pi)$-convergent. This question (1b) doesnt say anything but the text I wrote in this thread. I guess it's $(-pi,pi)$.. – kamikazi Sep 28 '16 at 18:41
  • @πr8 Yes, that was a typo! I'm sorry. I just corrected it! – kamikazi Sep 28 '16 at 18:42
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    I think the point was that your summation index is $i$, it should probably be $n$. And the function should probably be $e^{inx}$, not $e^{ix}$ independent of $n$. – Daniel Fischer Sep 28 '16 at 18:43
  • Okay, $L^2([-\pi,\pi])$ is basically equivalent to $L^2([0,2\pi])$. The point is that the functions are mutually orthogonal in that space. – Daniel Fischer Sep 28 '16 at 18:45
  • @DanielFischer Yes I switched it! Sorry once again. Im 100% that the function is e^ix – kamikazi Sep 28 '16 at 18:48
  • If the function is independent of $n$, $L^2$ is pretty much a red herring. Then the question is for which $p$ the series $\sum n^{-p}$ converges. – Daniel Fischer Sep 28 '16 at 18:57
  • @DanielFischer Sure, but why would the answer be p>1/2 ? – kamikazi Sep 28 '16 at 19:20
  • It wouldn't. It would be $p > 1$ then. The answer is $p > \frac{1}{2}$ for $$\sum_{n = 1}^{\infty} \frac{1}{n^p}e^{inx},$$ since by the orthogonality you have $$\Biggl\lVert \sum_{n = k}^m \frac{1}{n^p} e^{inx}\Biggr\rVert^2 = \sum_{n = k}^m \frac{1}{n^{2p}} \lVert e^{inx}\rVert^2.$$ – Daniel Fischer Sep 28 '16 at 19:24

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