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Why mpmath computes $\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right)$
Working with precision 500 decimal digits, mpmath in sage computes:
$$\sum_{n=2}^\infty (-1)^n\log(n)=\log\left(\frac1 2 \sqrt{2} \sqrt{\pi}\right).\tag{1}\label{1}$$
We believe the LHS of \eqref{1} ...
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