A summation method for this... $$ F(s) = -\sum_{n=2}^\infty \frac{(-1)^n}{n^s} = -(2^{1-s}-1)\zeta(s) \qquad\text{for $s>0$} $$ Differentiate: $$ -\sum_{n=2}^\infty \frac{(-1)^n\log n}{n^s} = F'(s) = 2^{1-s}(\log 2)\zeta(s) - (2^{1-s}-1)\zeta'(s) \qquad\text{for $s>0$} $$
Now the term-by-term limit as $s \to 0^+$ would be $$ \sum_{n=2}^\infty (-1)^n\log n $$ which diverges, but $$ \lim_{s\to 0^+} \big(2^{1-s}(\log 2)\zeta(s) - (2^{1-s}-1)\zeta'(s)\big) = 2(\log 2)\zeta(0) -\zeta'(0) \\ = -\log 2 + \frac{\log(2\pi)}{2} = \frac{\log \pi - \log 2}{2} $$