A cute limit $\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$ I'm sure that for many of you this is a limit pretty easy to compute, but my concern here
is a bit different, and I'd like to know if I can nicely compute it without using special functions. Do you have in mind such ways?
$$\lim_{m\to\infty}\left(\left(\sum_{n=1}^{m}\frac{1}{n}\sum_{k=1}^{n-1}\frac{(-1)^k}{k}\right)+\log(2)H_m\right)$$

$$
\begin{align}
\lim_{m\to\infty}\left(\sum_{n=1}^m\sum_{k=1}^{n-1}\frac{(-1)^k}{nk}+H_m\log(2)\right)
&=\lim_{m\to\infty}\sum_{n=1}^m\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\
&=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac{(-1)^{k-1}}{nk}\\
&=\sum_{k=1}^\infty\sum_{n=1}^k\frac{(-1)^{k-1}}{nk}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{H_k}k\\
&=\frac12\zeta(2)-\frac12\log(2)^2
\end{align}
$$
the last step is from this answer where only series manipulations are used, no special functions.
