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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.

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Partial fractions decomposition give us \begin{align*} \frac1{x^2(x+1)^2}&=-\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x+1}+\frac{1}{(x+1)^2} \end{align*}

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 $$\sum_{k=1}^n\frac1{k^2(k+1)^2}=2\sum_{k=1}^n\frac1{k^2}+\frac{2n+3}{(n+1)^2}-3$$

Then \begin{align*} \sum_{k=1}^{\infty}\frac1{k^2(k+1)^2}&=2\sum_{k=1}^{\infty}\frac1{k^2}+\lim_{n\to\infty}\left[\frac{2n+3}{(n+1)^2}-3\right]\\[3pt] &=2\cdot\frac{\pi^2}6+(0-3)\\ \sum_{k=1}^{\infty}\frac1{k^2(k+1)^2}&=\boxed{\color{blue}{\frac{\pi^2}3-3}} \end{align*}

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This is too long for a comment. It has been added for your curiosity.

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 $$\frac{\pi ^2}{3}-3-\frac 1{3000}\approx 0.289535$$

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