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Is there any way to simplify following summation?

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

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Consider \begin{align} \frac{1}{k^2 (k+1)^2} = \left(\frac{1}{k} - \frac{1}{k+1}\right)^{2} = \frac{1}{k^2} + \frac{1}{(k+1)^2} - \frac{2}{k} + \frac{2}{k+1} \end{align} for which \begin{align} \sum_{k=1}^{n} \frac{1}{k^2 (k+1)^2} &= \sum_{k=1}^{n} \frac{1}{k^2} + \sum_{k=2}^{n+1} \frac{1}{k^2} + 2 \sum_{k=2}^{n+1} \frac{1}{k} - 2 \sum_{k=1}^{n} \frac{1}{k} \\ &= -1 - \frac{1}{(n+1)^2} + 2 \sum_{k=1}^{n+1} \frac{1}{k^2} + \frac{2}{n+1} - 2 \\ &= -2 - \left( \frac{1}{n+1} - 1\right)^2 + \sum_{k=1}^{n+1} \frac{2}{k^2}\\ &= - \frac{3n^2 + 4n +2}{(n+1)^2} + \sum_{k=1}^{n+1} \frac{2}{k^2} \\ &= \frac{\pi^2}{6} - \frac{3n^2 + 4n +2}{(n+1)^2} - \psi_{1}(n+2) \end{align} where $\psi_{1}(x)$ is the derivative of the digamma function.

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By induction it can be shown that

$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac {n}{n+1}$$

That might be a first step

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