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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)

This is the Monthly Problem 11939. The answer is
$$ S = \frac{1}{2}\,(1 + \gamma - \log(2\pi)) + \frac{\pi^2}{72},$$
where $\gamma$ is the Euler-Mascheroni constant.

My submitted solution is long. Here is a short one.