I couldn't find any information online on how Euler calculated this to 16 digits of accuracy: $$ \gamma = - \ln{n} + \sum_{k=1}^\infty \frac1k $$ Obviously not like that, can somebody help please?
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3Euler calculated his $\gamma$ constant upto $16$ digits as he was Euler – Vanessa Oct 01 '22 at 17:27
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3It could have been through series acceleration, but I imagine he numerically evaluated an integral representation. By the way, this question would be a good fit for the HSM Stack Exchange. – J.G. Oct 01 '22 at 17:31
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@J.G. Thanks, didn't know it existed – Betydlig Oct 07 '22 at 19:18
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Some googling turns up this paper in which the authors claim that Euler did this computation using Euler-Maclaurin
$$H_n - \log n \approx \gamma + \frac{1}{2n} - \sum_{k \ge 1} \frac{B_{2k}}{2k} \frac{1}{n^{2k}}$$
where $B_{2k}$ are the Bernoulli numbers. Specifically the authors claim Euler went up to the $k = 6$ term of this expansion and calculated everything to $16$ digits of accuracy for $n = 10$.
Qiaochu Yuan
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1Ah, so it was series acceleration (I meant to include that link in my other comment) after all. +1. – J.G. Oct 01 '22 at 17:39