While working on a harder double sum, I (erroneously) reduced it to the sum below, which I recognized numerically to rapidly converge to $\log 2$.
Prove $$\lim_{m\to\infty}\sum_{k=1}^m\frac{2^{-k}}{k} = \log 2$$
The cute observation is that if you replace the $2^{-k}$ with $(-1)^{-k}$ you get the same result (only now the convergence becomes conditional).
Highlights:
For $\;|x|<1\;$ :
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies -\log(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}=\sum_{n=1}^\infty\frac{x^n}n$$
and now substitute $\;x=\frac12\;$ ...
By comparison with the geometric series with $1/2$ as common ratio, the series in the question converges absolutely, so we can write
\begin{align} & \sum_{k=1}^\infty \frac{2^{-k}}{k} \\ =& \sum_{k=1}^\infty 2^{-k} \int_0^1 x^{k-1} dx \\ =& \int_0^1 \sum_{k=1}^\infty 2^{-k} x^{k-1} dx \tag{Fubini's Theorem} \\ =& \frac12 \int_0^1 \sum_{k=1}^\infty \left(\frac{x}{2}\right)^{k-1} dx \\ =& \frac12 \int_0^1 \frac{1}{1-x/2} dx \\ =& \left[ -\ln\left|1-\frac{x}{2}\right| \right]_0^1 \\ =& \ln2. \end{align}
