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} diverges, so this isn't true.

Q1. Are there theoretical reasons for mpmath to compute \eqref{1}?

online code

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} diverges, so this isn't true.

Q1. Are there theoretical reasons for mpmath to compute \eqref{1}?

online code

Added Despite the interesting answers, I am ready bet mpmath doesn't do any analytic stuff not related to summation, it works purely numerically and the function is treated as black box, returning real number.

Working with precision 500 decimal digits, mpmath in sage computes:

$$\sum_{n=2}^\infty (-1)^n\log(n)=\log(1/2 \sqrt{2} \sqrt{\pi}) \qquad (1)$$

We believe the LHS of (1) diverges, so this isn't true.

Q1 Are there theoretical reasons mpmath to compute (1)?

online code

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} diverges, so this isn't true.

Q1. Are there theoretical reasons for mpmath to compute \eqref{1}?

online code