How did Euler calculate his $\gamma$ constant to 16 digits?

Question

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?

Answer 1

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 $n = 10$.