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I was doing a bit of self-study of sequences, and I considered $$\sum_{n=1}^{\infty}\frac {(-1)^n \ln(n)}{n} $$

which I then found out is ${\eta}'(1)$, the derivative of the Dirichlet Eta Function at $s=1$. I did a little bit of searching, and found a couple sources stating this equality: $${\eta}'(s)=2^{1-s}\ln(2)\zeta(s) + (1-2^{1-s})\zeta'(s)$$

But when I try to evaluate the sum by plugging 1 into the equation, I run into a problem, as $\zeta(1)=\infty$ and $\zeta'(1)=-\infty$, so I've got an indeterminate form here. I know that this series converges, and that $\eta'(1)=\gamma\ln2-\frac12 \ln^2(2)$.

My question is, how would I get to that value? What expression(s) could I use to prove that $\eta'(1)$ does actually equal that?

Gerry Myerson
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1 Answers1

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This is just a matter of finding the Laurent expansions of the functions involved in the formula

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

around $s = 1$ and then after multiplying them out some terms will cancel out and you'll be able to take the limit (or evaluate at $s = 1$) the resulting expression.

The first terms of the Laurent expansions are the following:

$$ \begin{eqnarray} 2^{1-s} = 1 - \log{2}(s-1) + \frac{1}{2}(\log{2})^2 (s - 1)^2 + O(s-1)^3\\ 1 - 2^{1-s} = \log{2}(s-1) - \frac{1}{2}(\log{2})^2 (s - 1)^2 + O(s-1)^3\\ \zeta(s) = \frac{1}{s-1} + \gamma + O(s-1)\\ \zeta'(s) = -\frac{1}{(s-1)^2} - \gamma_1 + \gamma_2 (s -1) -\frac{1}{2}\gamma_3 (s - 1)^2 + O(s - 1)^3 \end{eqnarray} $$

Then by multiplying them out carefully one gets

$$ \begin{eqnarray} &\eta'(s) = 2^{1-s} \log{2}\zeta(s) + (1 - 2^{1-s})\zeta'(s)\\ = &\left( \frac{\log{2}}{s-1} - (\log{2})^2 + \gamma\log{2} + O(s-1) \right) + \left( -\frac{\log{2}}{s-1} + \frac{1}{2}(\log{2})^2 + O(s-1) \right)\\ = &\gamma\log{2} -\frac{1}{2}(\log{2})^2 + O(s-1) \end{eqnarray} $$

so that

$$ \eta'(s) = \gamma\log{2} -\frac{1}{2}(\log{2})^2 + O(s-1) $$

Then from this last expression you get precisely that $\eta'(1) = \gamma\log{2} -\frac{1}{2}(\log{2})^2$ as you wanted. Note that I used the first terms of the Laurent series of the Riemann Zeta Function to get the expansion of $\zeta'(s)$.

Adrián Barquero
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