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 Post subject: Harmonic infinite sum
PostPosted: Sat May 06, 2017 7:56 pm 

Joined: Sat Nov 14, 2015 6:32 am
Posts: 146
Location: Melbourne, Australia
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)

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$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$


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PostPosted: Wed May 10, 2017 7:40 pm 

Joined: Tue May 10, 2016 3:56 pm
Posts: 32
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:

http://www.mat.uniroma2.it/~tauraso/AMM/AMM11939.pdf


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