Who's up to the challenge? 40

We see that $\sum_{n=1}^\infty \frac{(-1)^n}{n^s} \; = \; -\sum_{n=1}^\infty \frac{1}{n^s} + 2\sum_{n=1}^\infty \frac{1}{(2n)^s} \; =\; -\big(1 - 2^{1-s}\big)\zeta(s)$ and hence $\sum_{n=1}^\infty (-1)^n \frac{\ln n}{n^s} \; = \; \frac{d}{ds}\big[\big(1 - 2^{1-s}\big)\zeta(s)\big] \; =\; \big(1 - 2^{1-s}\big)\zeta'(s) + (\ln 2) 2^{1-s}\zeta(s)$ so that $\begin{array}{rcl} \displaystyle \sum_{n=1}^\infty (-1)^n \frac{\ln n}{n^2} & = & \displaystyle \tfrac12\big[\zeta'(2) + (\ln 2) \zeta(2)\big] \; = \; \tfrac{\zeta(2)}{2}\Big[\tfrac{\zeta'(2)}{\zeta(2)} + \ln 2\Big] \\ & = & \displaystyle \tfrac{\zeta(2)}{2}\big[ \gamma + \ln(2\pi) - 12\ln A + \ln2 \big] \; = \; \tfrac{\zeta(2)}{2}\big[\ln(4\pi) - 12\ln A + \gamma\big] \end{array}$ making the answer $2 + 2 + 4 + 12 + 1 =\boxed{21}$ .