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Double summation

Let \alpha>3 be a real number. Evaluate the series:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha}\]

Solution

Since the summands are all positive , we can sum by triangles. Thus,

    \begin{align*} \mathcal{S}&=\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{n}{\left ( n+m \right )^\alpha} \\ &=\sum_{n=2}^{\infty} \frac{\sum \limits_{k=1}^{n-1} k}{ n^\alpha} \\ &=\frac{1}{2} \sum_{n=2}^{\infty} \frac{n\left ( n-1 \right )}{n^\alpha} \\ &= \frac{1}{2} \left ( \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-2}} - \sum_{n=2}^{\infty} \frac{1}{n^{\alpha-1}} \right ) \\ &= \frac{\zeta\left ( \alpha-2 \right ) - \zeta\left ( \alpha-1 \right )}{2} \end{align*}

where \zeta is the Riemann zeta function.

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