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Consider the function $f:[-\pi, \pi) \to \mathbb{R}, f(x) = x(1+\cos(x))$, extended by $2\pi$-periodicity to the entire $\mathbb{R}$. Observe that its Fourier series of $f$ on $[-\pi, \pi)$ is $$\mathcal{F}\{f\}(x) = \frac{3}{2}\sin(x) + \sum_{n=2}^\infty \frac{2 \cdot (-1)^n}{(n-1)n(n+1)}\sin(nx). $$

How do we use the above Fourier series to compute $\displaystyle \sum_{n=2}^\infty \frac{(-1)^n}{(n-1)(n+1)} $ and $\displaystyle \sum_{n=2} ^\infty \frac{1}{(n-1)^2(n+1)^2}$?

I have no idea where to start. We know that the Fourier series of $f$ converges uniformly to $f$, but I cannot see what specific value of $x$ I should pick in order to eliminate the $n$ from the denominator of the coefficients. Parseval's identity doesn't help either.

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  • $\begingroup$ It’s seams you get wrong Fourier series. Look here and here. According to wolfram you forgot $\pi$ and there is no extra $n$ $\endgroup$ – Eugene Sirkiza Jan 9 at 21:54
  • $\begingroup$ You're right, I had a the wrong Fourier series (forgot a factor $2$), I edited it. However, there is no $\pi$, since the coefficients are divided by $\pi$. $\endgroup$ – aa33398 Jan 9 at 22:00
  • $\begingroup$ Yeah. I forgot about dividing on $\pi$. Now it seems correct. $\endgroup$ – Eugene Sirkiza Jan 9 at 22:04
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You can't really use $f(x)$ to compute the two series, because you cannot eliminate $n$ from the denominator, as you observed.

However, you can compute the Fourier series of $f'(x) = 1+\cos(x)-x\sin(x)$, defined on $[-\pi, \pi)$ (and extending it using periodicity).

The Fourier series of $f'(x)$ will be equal to $$f'(x) \sim \frac{3}{2}\cos(x) + \sum_{n=2}^\infty \frac{2 \cdot (-1)^n}{(n-1)(n+1)}\cos(nx), $$ which converges uniformly to $f'$. Now, observe that $$\sum_{n=2}^\infty \frac{(-1)^n}{(n-1)(n+1)} =\frac{1}{2} \left(f'(0)- \frac{3}{2}\right) $$ and you can compute $\displaystyle \sum_{n=2}^\infty \frac{1}{(n-1)^2(n+1)^2} $ using Parseval's identity for the Fourier series of $f'$.

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