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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
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  • $\begingroup$ Surely, just as the infinite sum of rationals need not be rational, there's no reason to expect an infinite sum of transcendentals to be transcendental? (Indeed, even a finite sum of transcendentals need not be transcendental!) $\endgroup$
    – LSpice
    Commented Oct 11, 2021 at 16:56

1 Answer 1

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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 $$ f'(1)= -\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n-1}\ln \left( n \right) }{n}}= \gamma \log 2-\frac{\log ^22}{2}. $$

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  • $\begingroup$ Thanks. How an infinite linear combination of logarithms and sqrt(2) can be rational? $\endgroup$
    – joro
    Commented Aug 27, 2013 at 12:27
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    $\begingroup$ 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? $\endgroup$
    – James Cranch
    Commented Aug 27, 2013 at 13:42
  • $\begingroup$ @JamesCranch linear combination of $e$ appears quite different to me from $\log{n}$. $\endgroup$
    – joro
    Commented Aug 27, 2013 at 16:05
  • $\begingroup$ Andrew, I fail to see which part of the question you answered. Maybe you answered "it is trivial" since you and Maple proved it. I just added "reference request" and "Was this known to Euler?". $\endgroup$
    – joro
    Commented Aug 27, 2013 at 16:50

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