Harmonic infinite sum
Harmonic infinite 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)
\[\mathcal{S} = \sum_{n=1}^{\infty} \left ( \mathcal{H}_n - \log n - \gamma - \frac{1}{2n} + \frac{1}{12n^2} \right )\]
(M.Omarjee)
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Harmonic infinite sum
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" onclick="window.open(this.href);return false;
$$ S = \frac{1}{2}\,(1 + \gamma - \log(2\pi)) + \frac{\pi^2}{72},$$
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" onclick="window.open(this.href);return false;
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