I've seen a similar question to this on this site: Calculate sum of $\sum_{n=1}^{\infty}(-1)^n\frac{\ln n}{n}$.
However, this sum is a little different because the argument within the natural log function does not match the denominator and the sum starts at n = 2 instead of 1.
When I plugged this into Wolfram Alpha I get a complex number as a result:
$$ \sum_{n=2}^{\infty}(-1)^n \frac{\ln(n-1)}{n} ≈-0.0834085 - 4.53932×10^-14 i$$
I can't tell if this is a April fool's joke by Wolfram or if this answer is legitimate. How does a complex number arise when all terms in the sum are real?
$$S=\sum_{n\geq 1}(-1)^{n+1}\frac{\log n}{n+1} $$ is conditionally convergent by Leibniz' test. By Frullani's theorem $\log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx$, hence
$$ S = \int_{0}^{1}\frac{u-u^2+u^2\log(2)-\log(1+u)}{u^2 \log u}\,du\approx -0.0966 $$ and in terms of the derivatives of the $\eta(s)=\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} $ function we have $$ S = -\eta'(1)+\eta'(2)-\eta'(3)+\eta'(4)-\ldots $$ where $-\eta'(1)=\frac{1}{2}\log^2(2)-\gamma\log(2)$ and the RHS is an absolutely convergent series, since $\eta'(s)$ has an exponential decay on $(1,+\infty)$. By the relation between the $\zeta$ and $\eta$ functions we have
$$ S = \frac{1}{2}\log^2(2)-\gamma\log(2)+2\log(2)-2\log^2(2)+\sum_{n\geq 2}(-1)^n\left(1-\frac{2}{2^n}\right)\zeta'(n) $$ or $$ \boxed{S = -\frac{3}{2}\log^2(2)+(2-\gamma)\log(2)-\sum_{n\geq 1}\frac{\log(n)}{(n+1)(2n+1)}}$$ which allows an accurate numerical evaluation: $S\approx -0.096614934732226887$.