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
$$
\lim_{s\to 0^+} \big(2^{1-s}(\log 2)\zeta(s) - (2^{1-s}-1)\zeta'(s)\big)
= 2(\log 2)\zeta(0) -\zeta'(0)
\\
= -\log 2 + \frac{\log(2\pi)}{2} = \frac{\log \pi - \log 2}{2}
$$