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Compute the value of the following double sum:

$$\sum_{\mu=1}^n\sum_{\upsilon=\mu+1}^n\frac{\mu^2}{\upsilon(2\upsilon-1)}$$

I started by simply trying to compute the value of the inner sum:

$$\sum_{\upsilon=\mu+1}^n\frac{\mu^2}{\upsilon(2\upsilon-1)}$$

treating $\mu$ as a generic constant. I can pull it out and do partial fraction decomposition to see if that helps:

$$\mu^2\sum_{\upsilon=\mu+1}^n\left[\frac{2}{2\upsilon-1}-\frac{1}{\upsilon}\right]$$

But this doesn't really seem to get me very far. I feel pretty stuck at this point. How would you evaluate this sum?

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By exchanging the two sums we get $$\begin{align} \sum_{\mu=1}^n\sum_{\upsilon=\mu+1}^n\frac{\mu^2}{\upsilon(2\upsilon-1)}&=\sum_{\upsilon=2}^n\frac{1}{\upsilon(2\upsilon-1)}\sum_{\mu=1}^{\nu-1}\mu^2\\ &=\frac{1}{6}\sum_{\upsilon=2}^n\frac{\nu(\nu-1)(2\upsilon-1)}{\upsilon(2\upsilon-1)}\\ &=\frac{1}{6}\sum_{\upsilon=2}^n( \nu-1)=\frac{n(n-1)}{12}\end{align}$$ where we used Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ and then Proof for formula for sum of sequence $1+2+3+\ldots+n$?

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