Scary Limit.

We see that $\begin{array}{rcl} \displaystyle \sum_{n=1}^m \frac{1}{n} \left(\sum_{k=1}^{n-1} \frac{(-1)^k}{k}\right) - \ln(2)\, H_m & = & \displaystyle\sum_{n=1}^m \frac{1}{n}\left(\sum_{k=1}^{n-1} \frac{(-1)^k}{k} - \ln 2\right) \\ & = & \displaystyle \sum_{n=1}^m \frac{1}{n}\left(\sum_{k=n}^\infty \frac{(-1)^{k-1}}{k}\right) \; =\; \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\left(\sum_{n=1}^{\mathrm{min}(k,m)} \frac{1}{n}\right) \\ & = & \displaystyle \sum_{k=1}^m \frac{(-1)^{k-1}}{k}H_k + \left(\sum_{k=m+1}^\infty \frac{(-1)^{k-1}}{k}\right)H_m \end{array}$ Now the second term is $O\big(\tfrac{\ln m}{m}\big)$ as $m \to \infty$ , and hence $\lim_{m\to\infty} \sum_{n=1}^m \frac{1}{n} \left(\sum_{k=1}^{n-1} \frac{(-1)^k}{k}\right) - \ln(2) \,H_m \;= \; \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}H_k$ which I proved elsewhere to be equal to $\tfrac12\zeta(2) - \tfrac12(\ln 2)^2 \; = \; \tfrac{1}{12}\pi^2 - \tfrac12(\ln 2)^2$ which makes the answer $2 + 12 + 2 + 2 = 18$ .