I am trying to compute a sum given the Fourier series of the function $f(x)=x(1+\cos x)$, which is
$$\frac{3}{2} \sin x +2\sum_{n=2}^{\infty}(-1)^n\frac{\sin nx}{(n-1)n(n+1)}$$
I have found that this series converges uniformly to $f$ on $(-\pi,\pi)$, but now that I have to compute the value of $$\sum_{n=2}^{\infty}\frac{(-1)^n}{(n-1)(n+1)}$$ I am stuck.
Could someone please give me a hint so I can solve it?
Kind regards
It is not too hard to show that $f$ is Lipschitz continuous as a periodic function on $[-\pi,\pi]$. Hence the Fourier series converges everywhere.
Hence we have $f(x) = {3 \over 2} \sin x + 2 \sum_{n=1}^\infty (-1)^n {\sin nx \over (n-1)n(n+1)}$ as long as $|x| \le \pi$.
It is straightforward to see that the derivatives of the summand $(-1)^n{\cos nx \over (n-1)(n+1)}$ are uniformly convergent everywhere, hence $f'(x) = {3 \over 2} \cos x + 2 \sum_{n=1}^\infty (-1)^n {\cos nx \over (n-1)(n+1)}$.
Evaluating at $x=0$ gives $f'(0) = {3 \over 2} + 2 \sum_{n=1}^\infty (-1)^n {1 \over (n-1)(n+1)}$.
Since $f'(x) = 1+\cos x + x \sin x$ for $|x| < \pi$, we see that $f'(0) = 2$ and from this we can compute the value of the summation.