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A harmonic sum

Let \mathcal{H}_n denote the n-th harmonic sum. Evaluate the sum:

    \[\mathcal{S} = \sum_{n=1}^{\infty} \left ( \mathcal{H}_n - \log n - \gamma - \frac{1}{2n} + \frac{1}{12n^2} \right )\]

(M. Omarjee)

Solution [Roberto Tauraso]

It is easy to see that

    \[\sum_{k=1}^{n} \mathcal{H}_k = (n+1) \mathcal{H}_n - n\]

since

    \begin{align*} \sum_{k=1}^{n} \mathcal{H}_k &=\sum_{k=1}^{n} \sum_{j=1}^{k} \frac{1}{j} \\ &= \sum_{j=1}^{n} \sum_{k=j}^{n} \frac{1}{j} \\ &= \sum_{j=1}^{n} \frac{n+1- j}{j}\\ &= \left ( n+1 \right ) \mathcal{H}_n - n \end{align*}

Hence

\begin{aligned} \sum_{k=1}^{n} \left ( \mathcal{H}_k - \log k - \gamma - \frac{1}{2k} \right ) &=\left ( n + 1 \right ) \mathcal{H}_n - n - \log n! - \\ & \quad \quad \quad -n \gamma - \frac{1}{2} \mathcal{H}_n \\ &= \left ( n + \frac{1}{2} \right ) \bigg [ \log n +\gamma + \frac{1}{2n} + \\ & \quad \quad \quad +\mathcal{O} \left ( \frac{1}{n^2} \right ) \bigg ] - n - n \gamma - \\ & \quad \;\;\; \bigg[ n \log n - n + \frac{\log 2 \pi}{2} + \\ & \quad \quad \quad + \frac{\log n}{2}+ \mathcal{O} \left ( \frac{1}{n} \right ) \bigg] \\ &=\frac{1 + \gamma}{2} - \frac{\log 2\pi}{2} + \mathcal{O} \left ( \frac{1}{n} \right ) \end{aligned}

Hence

    \[\mathcal{S} = \frac{1+\gamma }{2} - \frac{\log 2\pi}{2} + \frac{\pi^2}{72}\]

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