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Calculate the Fourier series of the function $f(x)=\sin{(ax)}$ on the interval $[-\pi,\pi]$ and then calculate: $$\sum_{n=1}^{\infty}{\frac{n^{2}}{(4n^2-1)^2}}$$

I've already calculated the Fourier series for this function:

$$\sin{(ax)}=\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{2n \sin{(a \pi)} (-1)^{n}}{a^{2}-n^{2}}\sin{(nx)}$$

I can't really find a suitable $a$ to sum the series in the problem. For one thing, the polynomial in the denominator of the series has a higher degree than in the Fourier series. I've tried using partial fractions but the degrees still differ.

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    $\begingroup$ I hate people who remove a question they ask. Please in the future think hard before asking the question and then post it on the site. I am sorry to tell you this over here(the wrong place in my opinion), but sadly you deleted your previous question leaving me no choice but to communicate this with you from here. For next time, please reconsider your actions. If you found the answer, then help others by posting it. Don't be selfish. There is no problem in answering your own question! $\endgroup$ – John Jan 13 '14 at 23:21
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Solution: Use Parseval's identity.

Set $f(x)=\sin\left(ax\right)$ with $a=\dfrac 1 2$, for all $x\in [-\pi, \pi]$. Find $b_n=\dfrac{(-1)^{n+1}8n}{\pi(4n^2-1)}$ for all $n\in \mathbb N$ and $\displaystyle \dfrac 2 \pi\int \limits_0^\pi\left(\sin\left(\dfrac x 2\right)\right)^2\mathrm dx=1$.

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  • $\begingroup$ Thanks for the suggestion. I used the identity and got: $$\sum_{n=-\infty}^{\infty} \frac{2i(a\cos{(\pi a}}\sin{\pi n}-n \sin{\pi a} \cos{\pi n}{a^{2}-n^{2}}=\pi - \frac{\sin{2 \pi a}}{2a}$$ Now I don't really know how to continue, though. $\endgroup$ – gndz Jan 12 '14 at 20:32
  • $\begingroup$ Sorry, here's the correct expression:$$\sum_{n=-\infty}^{\infty} \frac{2i(a \cos{(\pi a}\sin{\pi n}-n \sin{\pi a} \cos{\pi n}}{{a^{2}-n^{2}}}=\pi - \frac{\sin{2 \pi a}}{2a}$$ $\endgroup$ – gndz Jan 12 '14 at 20:39
  • $\begingroup$ This time for real...$$\sum_{n=-\infty}^{\infty} \frac{2i(a \cos{(\pi a}\sin{\pi n}-n \sin{\pi a} \cos{\pi n))}}{{a^{2}-n^{2}}}=\pi - \frac{\sin{2 \pi a}}{2a}$$ $\endgroup$ – gndz Jan 12 '14 at 20:39
  • $\begingroup$ @Mike I updated the link in the answer and added some details. $\endgroup$ – Git Gud Jan 12 '14 at 21:29
  • $\begingroup$ OK I think I'm starting to understand, kind of. So if I understand correctly the identity states that the sum of the square of the coefficients of the Fourier series is equal to that easily computable integral? But I still don't know how to find the negative coefficients. $\endgroup$ – gndz Jan 12 '14 at 21:35

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