# Series with denominator that is square of product of consecutive numbers

Evaluate $\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}$.

I've tried to use a telescoping approach, but it doesn't seem to work. It doesn't also seem to converge to a "pretty" value as well.

Then \begin{align*} \sum_{k=1}^n\frac1{k^2(k+1)^2}&=\sum_{k=1}^n\left(\frac2{k+1}-\frac2k\right)+\sum_{k=1}^n\left[\frac1{k^2}+\frac1{(k+1)^2}\right]\\[3pt] &=\frac2{n+1}-2+\frac1{(n+1)^2}+2\sum_{k=1}^n\frac1{k^2}-1 \end{align*} Where we have used the fact that the first sum on the RHS is telescopic. So $$\sum_{k=1}^n\frac1{k^2(k+1)^2}=2\sum_{k=1}^n\frac1{k^2}+\frac{2n+3}{(n+1)^2}-3$$
Ángel Mario Gallegos having given the good solution to the problem, let me address the problem of the partial sums starting from what Ángel Mario Gallegos wrote $$S_n=\sum_{k=1}^n\frac1{k^2(k+1)^2}=2\sum_{k=1}^n\frac1{k^2}+\frac{2n+3}{(n+1)^2}-3$$ But, $$\sum_{k=1}^n\frac1{k^2}=H_n^{(2)}$$ where appear the generalized harmonic numbers. This makes $$S_n=2H_n^{(2)}+\frac{2n+3}{(n+1)^2}-3$$ For large values of $n$, we can use the asymptotics $$H_n^{(a)}=n^{-a} \left(-\frac{n}{a-1}+\frac{1}{2}-\frac{a}{12 n}+\frac{a^3+3 a^2+2 a}{720 n^3}+O\left(\frac{1}{n^4}\right)\right)+\zeta (a)$$ which, in the present case, reduces to $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6 n^3}+O\left(\frac{1}{n^4}\right)$$ Continuing the expansion for the other terms, we should get $$S_n=\left(\frac{\pi ^2}{3}-3\right)-\frac{1}{3 n^3}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Using $n=10$, we should find $$S_{10}=\frac{22254209}{76839840}\approx 0.289618$$ while the approximation would give $$\frac{\pi ^2}{3}-3-\frac 1{3000}\approx 0.289535$$