Calculate the given sum

Question

Problem: $$\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}$$

This is a problem from a course of Fourier Series and the only hint was to use partial fraction decomposition.

I'm sure that this is meant to be solved by using Plancherel's formula: $$ \| f\|_2^2= \sum{k \in \mathbb{Z} } |g_n|^2 = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n^2 + b_n^2)$$ where $g_n$ is the n-th coefficient in an exponential fourier series of a function $f\in L^2([0,1])$, $g_n = \int_{0}^{1}f(x)e^{-2\pi inx}dx$, $a_n$ and $b_n$ are standard coefficients in a trigonometric Fourier series of a function $f\in L^2([-\pi,\pi]$).

Answer 1

As pointed out in the comments, we have $$\frac{1}{k^2(k+1)^2}=\frac{1}{k^2}+\frac{1}{(k+1)^2}+\frac{2}{k+1}-\frac{2}{k}.$$ Now, a standard sum is $$\sum_{k=1}^\infty\frac{1}{k^2}=\pi^2/6$$ so $\sum_{k=1}^\infty1/(k+1)^2=\pi^2/6-1$ so both these sums add up to $\pi^2/3-1$. The remaining part is $$2\sum_{k=1}^\infty\left(\frac{1}{k+1}-\frac{1}{k}\right)=2(1/2-1+1/3-1/2+1/4-1/3+1/5-1/4+1/6-1/5+\ldots)=2\cdot-1=-2$$ so the final answer is $$\pi^2/3-3.$$

Answer 2

Problem: $$\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}=\sum_{k=1}^{\infty}\frac1{k^2}+\sum_{k=1}^{\infty}\frac1{(k+1)^2}-2\sum_{k=1}^{\infty}{\left(\frac1k-\frac1{k+1}\right)}$$ $$=\frac{\pi^2}{6}+\frac{\pi^2}{6}-1-2=\frac{\pi^2}{3}-3$$