I first computed the inverse Discrete time Fourier transform of H:
$$h(n)=\frac{1}{2\pi} \int_{-\pi}^{\pi} iwe^{iwn}dw$$
Using integration by parts, I got $\frac{-1}{n}$
Now, using Parseval's theorem:
$$\sum_{-\infty}^{\infty} (\frac{1}{-n})^2= \frac{1}{2\pi} \int_{-\pi}^{\pi} (iw)^2dw$$
=$$\frac{1}{2\pi}[\frac{-w^3}{3}]_{-\pi}^{\pi}$$
=$$-\frac{\pi^2}{3}$$
This seems wrong to me because the sum is supposed to be positive while this is negative.
Also, this sum from $-\infty$ to $\infty$ is going through a singularity at 0. How can it be finite? Please explain.
And even after that, I need the sum from 1 to $\infty$ instead of this. I could maybe half this calculated sum to get that required value if there was no singularity.
First, you can compute $h(0)$ separately which gives you $0$, because $(\omega \mapsto i\omega)$ is an odd function.
And notice that in Parseval identity you have the integral of the square of the modulus of the function : $\int_{-\pi}^{\pi} |iw|^2\mbox{d}\omega$, which removes the extra minus sign you had.
Therefore you have \begin{align*} \sum_{n \leq -1} \left(\dfrac{(-1)^n}{n}\right)^2 + 0 + \sum_{n \geq 1} \left(\dfrac{(-1)^n}{n}\right)^2 &= \dfrac{1}{2\pi}\int_{-\pi}^{\pi} |iw|^2\mbox{d}\omega \\ 2\sum_{n \geq 1} \dfrac{1}{n^2}&=\dfrac{\pi^2}{3} \\ \sum_{n \geq 1} \dfrac{1}{n^2}&=\dfrac{\pi^2}{6}, \end{align*} which is the well-known result.