1. ## series

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \ \frac{\ln n}{n}$

The sum should be $\displaystyle \gamma \ln 2 - \frac{(\ln 2)^{2}}{2}$ .

2. Remembering the definition of the 'Dirichlet eta function'...

$\displaystyle \eta(s)= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}$ (1)

... it is easy to see that is...

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n}= \eta^{'} (1)$ (2)

Not so easy however is the numerical computation of (2)...

Kind regards

$\chi$ $\sigma$

3. 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

$\displaystyle \eta(s)= - \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{s}}= (1-2^{1-s})\ \zeta (s)$ (1)

... where $\zeta(s)$ is the so called 'Riemann zeta function'. The goal is to find a series expansion 'somewehre around' $s=1$ of its derivative...

$\displaystyle \eta^{'}(s)= \ln 2\ 2^{1-s}\ \zeta(s) + (1-2^{1-s})\ \zeta^{'}(s)$ (2)

At this scope we use the [well known] Laurent expansion...

$\displaystyle \zeta(s) = -\frac{1}{1-s} + \gamma + \sum_{n=1}^{\infty} \gamma_{n}\ \frac{(1-s)^{n}}{n!}$ (3)

... where $\gamma= .577215665...$ is the Euler's constant and the $\gamma_{n}$ are the so called 'Stieltjes constants'. Deriving (3) we obtain...

$\displaystyle \zeta^{'}(s) = -\frac{1}{(1-s)^{2}} + \sum_{n=1}^{\infty} (-1)^{n}\ n\ \gamma_{n}\ \frac{(1-s)^{n-1}}{n!}$ (4)

... and is...

$\displaystyle 2^{1-s}= \sum_{n=0}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!} \implies 1-2^{1-s}= - \sum_{n=1}^{\infty} \ln^{n} 2\ \frac{(1-s)^{n}}{n!}$ (5)

... so that we have all the terms we need to compute (2). We are interested to the value of (2) in $s=1$, so that we stop the computation at the constant term obtaining...

$\displaystyle \eta^{'} (s)= -\frac{\ln 2}{1-s} + \gamma\ \ln 2 - \ln^{2} 2 +...+\frac{\ln 2}{1-s} + \frac{\ln^{2} 2}{2}+...$ (6)

... and finally...

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n} \frac{\ln n}{n} = \eta^{'} (1) = \gamma\ \ln 2 - \frac{\ln^{2} 2}{2}$ (7)

Kind regards

$\chi$ $\sigma$