TITLE: Prove $\lim_{m\to\infty}\sum_{k=1}^m\frac{2^{-k}}{k} = \log 2$
QUESTION [7 upvotes]: 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).

REPLY [5 votes]: 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}