# Compute $\sum_{n=2}^\infty \frac{(-1)^n}{(n-1)(n+1)}$ and $\sum_{n=2} ^\infty \frac{1}{(n-1)^2(n+1)^2}$ using Fourier series

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.

• It’s seams you get wrong Fourier series. Look here and here. According to wolfram you forgot $\pi$ and there is no extra $n$ – Eugene Sirkiza Jan 9 at 21:54
• 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 $\pi$. – aa33398 Jan 9 at 22:00
• Yeah. I forgot about dividing on $\pi$. Now it seems correct. – Eugene Sirkiza Jan 9 at 22:04

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 $$f'$$.