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Arithmotheoretic sum

Evaluate the sum

    \[\mathcal{S} = \sum_{n=1}^\infty\frac{1}{n}\sum_{d\mid n} \frac{d}{n+d^2}\]

Solution

The sum converges absolutely , so we can switch the order of summation; hence:

    \begin{align*} \sum_{n=1}^{\infty} \frac{1}{n} \sum_{d\mid n} \frac{d}{n+d^2} &= \sum_{d=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{dk} \frac{d}{dk+d^2} \\ &= \sum_{d=1}^{\infty} \frac{1}{d} \sum_{k=1}^{\infty} \frac{1}{k(d+k)} \\ &= \sum_{d=1}^{\infty} \frac{1}{d^2} \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{d+k} \right) \\ &= \sum_{d=1}^{\infty} \frac{\mathcal{H}_d}{d^2} \end{align*}

The last sum equals 2\zeta(3) and hence

    \[\mathcal{S} = 2\zeta(3)\]

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