TITLE: Alternating series: $\sum\limits_{n= 1}^{\infty} (-1)^{n-1} \frac{\ln(n)}{n}$ convergence?
QUESTION [5 upvotes]: Determine whether the series converges absolutely, conditionally or diverges?
$$\sum\limits_{n= 1}^{\infty} (-1)^{n-1} \frac{\ln(n)}{n}$$
I know that $\sum|a_{n}|$ diverges by using the comparison test:
$$\frac{\ln(n)}{n} > \frac{1}{n}$$ and the smaller, r.h.s being the divergent harmonic series.
So, should my conclusion for the alternating series be divergent or convergent conditionally*?

* How to estimate whether the alternating series terms are cancelling?

REPLY [4 votes]: Since $\frac{\log n}{n}$ is a decreasing function on $n\geq 3$ and $\{(-1)^n\}_{n\geq 1}$ is a sequence with bounded partial sums, the series is convergent by Dirichlet's test. Moreover, by exploiting the integral representation for $\log(n)$ provided by Frullani's theorem, we have:
$$ S=\sum_{n\geq 1}\frac{(-1)^{n-1}\log(n)}{n}=\int_{0}^{+\infty}\frac{e^{-x}\log(2)-\log(1+e^{-x})}{x}\,dx\tag{1}$$
and by the integral representation for the Euler-Mascheroni constant we have:
$$ \sum_{n\geq 1}\frac{(-1)^{n-1}\log(n)}{n} = \color{red}{\frac{\log^2(2)}{2}-\gamma\log(2)}\approx -0.1598689 .\tag{2}$$