# Is this known alternating sum for Euler's constant?

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?

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


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 $$f'(1)= -\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}\ln \left( n \right) }{n}}= \gamma \log 2-\frac{\log ^22}{2}.$$
• Same way that the infinite linear combination $$\frac{1}{e}+\frac{1}{e}+\frac{1}{2e}+\frac{1}{6e}+\frac{1}{24e}+\cdots$$ can be 1? – James Cranch Aug 27 '13 at 13:42
• @JamesCranch linear combination of $e$ appears quite different to me from $\log{n}$. – joro Aug 27 '13 at 16:05