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Could you please help me solve this problem:

I need use Fourier series of $f(x)=x^2+x$ ,$x\in(-\pi,\pi)$

to prove that $\sum_{n\ge1} \frac 1{n^2}= \frac{\pi^2}6$.

I calculated the Fourier series: $x^2+x=\frac {\pi^2}3+\sum_{n\ge1} \frac 4{n^2}(-1)^ncosnx-\frac 2n (-1)^n sinnx$.

And could not find any $-\pi<x<\pi$ that could solve my problem.

I double-checked my calculations and could not find any problem with the series, could you please give me some hint about my error ?

I know that we can calculate $\sum_{n\ge1} \frac 1{n^2}= \frac{\pi^2}6$ by using the Fourier expansion of $f(x)=x$ but I need to prove this using different function.

Thanks.

user97484
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  • Agreed, i also can't find any proper value! – Someone May 13 '15 at 14:12
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    If you know how the Fourier series behaves at jump discontinuities of the function, you can use $x = \pi$ (or $x = -\pi$). Otherwise, you can use $x = 0$ and a little bit of series manipulation. – Daniel Fischer May 13 '15 at 14:15
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    Could you please explain a bit more about series manipulation that can bring us from $0=\frac{\pi^2}3 +4\sum_{n\ge1}\frac{(-1)^n}{n^2}$ to $\sum_{n\ge1} \frac 1{n^2}= \frac{\pi^2}6$ ? – user97484 May 13 '15 at 14:29
  • @DanielFischer , could you please illustrate on my behalf aswell in answer. I had seen similar question in my brother's book. And couldn't find a nice value back then about 7-8 months ago. I am kinda interested to see. Only if you'd like to :). – Someone May 13 '15 at 14:33
  • @Mann Sure thing, done. – Daniel Fischer May 13 '15 at 14:51

1 Answers1

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If we take $x = \pi$, the right hand side becomes

$$\frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4}{n^2},$$

which is very convenient to determine $\sum\limits_{n=1}^\infty \frac{1}{n^2}$ provided we know what the Fourier series converges to for $x = \pi$.

If $f$ has a jump discontinuity at $x_0$, and is otherwise well-behaved in a neighbourhood of $x_0$ - I can't remember the criteria off the top of my head, but Lipschitz continuity is sufficient - then the Fourier series converges to the arithmetic mean of the one-sided limits of $f$ at $x_0$.

Here, we have

$$\lim_{x\to \pi^-} f(x) = \pi^2 + \pi\quad \text{and} \quad \lim_{x\to \pi^+} f(x) = \pi^2 - \pi,$$

so the arithmetic mean is $\pi^2$, and

$$\pi^2 = \frac{\pi^2}{3} + \sum_{n=1}^\infty \frac{4}{n^2}$$

easily yields

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$

If we don't know the convergence behaviour of Fourier series at (nice enough) jump discontinuities, the best we can do is evaluate at $x = 0$. Then we obtain

$$0 = \frac{\pi^2}{3} + 4\sum_{n=1}^\infty \frac{(-1)^n}{n^2},$$

which we rearrange to

$$\frac{\pi^2}{12} = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = \sum_{n=1}^\infty \frac{1}{n^2} - 2\sum_{m=1}^\infty \frac{1}{(2m)^2} = \biggl(1 - \frac{2}{4}\biggr)\sum_{n=1}^\infty \frac{1}{n^2},$$

from which we again obtain

$$\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$$

Daniel Fischer
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