TITLE: Is this known alternating sum for Euler's constant?
QUESTION [5 upvotes]: This probably is known, but Wolfram Alpha doesn't recognize it
and couldn't find it in Mathworld (there is something close,
but using floor).
We have 
$\lim_{s \to 1} (\zeta(s)-1/(s-1)) = \gamma$
Also 
$F(s) = \zeta(s) = \frac{1}{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} $.
According to Maple 13:
$$\lim_{s \to 1} (F(s)-1/(s-1)) = \sum _{n=1}^{\infty }-{\frac { \left( -1 \right) ^{n-1}\ln  \left( n \right) }{n}} \left( \ln  \left( 2 \right)  \right) ^{-1}+1/2\,\ln \left( 2 \right) = \gamma \qquad (1) $$

Is (1) known and/or trivial?

I believe all terms and partial sums except the first of the sum are transcendental.

Intuitive explanation how (1) could be hypothetically rational?



Reference request? Was this known to Euler?

Numerically (1) is correct to precision at least $500$ decimal digits.
Sage code:
nsu=1/2*mpmath.log(2)-mpmath.nsum(lambda n:  (-1)**(n-1)*mpmath.log(n)/n ,[1, mpmath.inf])/ mpmath.log(2);nsu

REPLY [11 votes]: Denote 
$$
f(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}=(1-2^{1-s})\zeta(s). 
$$
Expanding into series and using $\zeta(s)=\frac1{s-1}+\gamma+O((s-1))$ leads to
$$
f(s)=\log 2+(s-1) \left(\gamma  \log 2-\frac{\log
   ^22}{2}\right)+O\left((s-1)^2\right).
$$
Differentiating both sides gives
$$
f'(1)=
-\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}\ln  \left( n \right) }{n}}= 
\gamma  \log 2-\frac{\log
   ^22}{2}.
$$