Limit of sequence for an alternating series Find the limit of the sequence $y_n = \sum_{k=1}^n(-1)^k\frac{\ln{k}}{k}$.
I have tested the limit for convergence using Lebiniz's test for alternating series and it is indeed convergent. I can't seem to figure out how to get to the answer, though.
 A: Using Euler-Maclaurin we can show existence of 
$$C= \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{\log k}{k} - \int_1^n\frac{\ln x}{x} \, dx\right) = \lim_{n \to \infty} \left(\sum_{k=1}^n \frac{\log k}{k} - \frac{(\ln n)^2}{2}\right)$$
Note that
$$\sum_{k=1}^{2n} (-1)^k \frac{\ln k}{k} = -\left(\sum_{k=1}^{2n}\frac{\ln k}{k} - \frac{(\ln(2n))^2}{2}\right) + 2\sum_{k=1}^n\frac{\ln(2k)}{2k} - \frac{(\ln(2n))^2}{2}$$
Using $\ln(2n) = \ln 2 + \ln  n$, the RHS reduces to 
$$\sum_{k=1}^{2n} (-1)^k \frac{\ln k}{k} = -\left(\sum_{k=1}^{2n}\frac{\ln k}{k} - \frac{(\ln(2n))^2}{2}\right) +\left(\sum_{k=1}^{n}\frac{\ln k}{k} - \frac{(\ln n)^2}{2}\right) \\+ \ln 2\left(\sum_{k=1}^n \frac{1}{k} - \ln n \right) - \frac{(\ln 2)^2}{2}$$
Taking the limit as $n \to \infty$ we get
$$\sum_{k=1}^{\infty} (-1)^k \frac{\ln k}{k} = -C + C + \gamma \ln 2 - \frac{(\ln 2)^2}{2}= \ln 2 \left(\gamma - \frac{\ln 2}{2} \right)$$
A: Here is another technique. Let $$f(s) =\sum_{n=1}^{\infty} (-1)^{n-1}\frac{1}{n^s}$$ then we have $$f(s) - \zeta(s) = - \frac{\zeta(s)}{2^{s-1}}$$ or $$f(s) =(1-2^{1-s})\zeta(s)$$ The desired limit in question is $f'(1)$. Differentiate the above with respect to $s$ and let $s\to 1$. This will give the desired result.

I later found the same approach given in this answer.

This answer justifies the appearance of $\gamma$ as a constant term in Laurent series for $\zeta(s) $ and thereby makes the approach in this answer non-circular (see comments below). 
