Problem with the Dirichlet Eta Function 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?
 A: 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 $\zeta'(s)$.
