$\displaystyle \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \ \frac{\ln n}{n} $
The sum should be $\displaystyle \displaystyle \gamma \ln 2 - \frac{(\ln 2)^{2}}{2} $ .
Remembering the definition of the 'Dirichlet eta function'...
$\displaystyle \displaystyle \eta(s)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}$ (1)
... it is easy to see that is...
$\displaystyle \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n}= \eta^{'} (1)$ (2)
Not so easy however is the numerical computation of (2)...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
I haven't thought about this deeply or anything, but this almost seems like it can be done using some kind of summation by parts argument.
Summation by parts - Wikipedia, the free encyclopedia
Let's start with the so called 'Dirichlet eta function'...
$\displaystyle \displaystyle \eta(s)= - \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{s}}= (1-2^{1-s})\ \zeta (s)$ (1)
... where $\displaystyle \zeta(s)$ is the so called 'Riemann zeta function'. The goal is to find a series expansion 'somewehre around' $\displaystyle s=1$ of its derivative...
$\displaystyle \displaystyle \eta^{'}(s)= \ln 2\ 2^{1-s}\ \zeta(s) + (1-2^{1-s})\ \zeta^{'}(s)$ (2)
At this scope we use the [well known] Laurent expansion...
$\displaystyle \displaystyle \zeta(s) = -\frac{1}{1-s} + \gamma + \sum_{n=1}^{\infty} \gamma_{n}\ \frac{(1-s)^{n}}{n!}$ (3)
... where $\displaystyle \gamma= .577215665...$ is the Euler's constant and the $\displaystyle \gamma_{n}$ are the so called 'Stieltjes constants'. Deriving (3) we obtain...
$\displaystyle \displaystyle \zeta^{'}(s) = -\frac{1}{(1-s)^{2}} + \sum_{n=1}^{\infty} (-1)^{n}\ n\ \gamma_{n}\ \frac{(1-s)^{n-1}}{n!}$ (4)
... and is...
$\displaystyle \displaystyle 2^{1-s}= \sum_{n=0}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!} \implies 1-2^{1-s}= - \sum_{n=1}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!} $ (5)
... so that we have all the terms we need to compute (2). We are interested to the value of (2) in $\displaystyle s=1$, so that we stop the computation at the constant term obtaining...
$\displaystyle \displaystyle \eta^{'} (s)= -\frac{\ln 2}{1-s} + \gamma\ \ln 2 - \ln^{2} 2 +...+\frac{\ln 2}{1-s} + \frac{\ln^{2} 2}{2}+...$ (6)
... and finally...
$\displaystyle \displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n} = \eta^{'} (1) = \gamma\ \ln 2 - \frac{\ln^{2} 2}{2}$ (7)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$