TITLE: Show that $\sum_{k=1}^{\infty}\frac{2^{-k}}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}$ without evaluating either sum
QUESTION [6 upvotes]: Show that
$$
\sum_{k=1}^{\infty}\frac{2^{-k}}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}
$$
without evaluating either sum.
This is inspired by
Prove $\lim_{m\to\infty}\sum_{k=1}^m\frac{2^{-k}}{k} = \log 2$.
This, of course,
follows from
$\ln(1+x)
=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}x^{k}}{k}
$
for
$-1 \lt x \le 1$.
But I am not allowing that!
What I am looking for
is a way to
manipulate one of the series
to convert it to
the other.

REPLY [2 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 1}^{\infty}{2^{-k} \over k} & =
\sum_{k = 1}^{\infty}2^{-k}\int_{0}^{1}t^{k - 1}\,\dd t =
\int_{0}^{1}\sum_{k = 1}^{\infty}\pars{t \over 2}^{k}\,{\dd t \over t} =
\int_{0}^{1}{t/2 \over 1 - t/2}\,{\dd t \over t}
\\[5mm] & =
\int_{0}^{1}{\dd t \over 2 - t} \,\,\,\stackrel{t\ \mapsto\ 1 - t}{=}\,\,\,
\int_{0}^{1}{\dd t \over 1 + t} =
\int_{0}^{1}\sum_{k = 0}^{\infty}\pars{-1}^{k}\,t^{k}\,\dd t =
\sum_{k = 0}^{\infty}\pars{-1}^{k}\int_{0}^{1}t^{k}\,\dd t
\\[5mm] & =
\sum_{k = 0}^{\infty}{\pars{-1}^{k} \over k + 1} =
\bbx{\sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k}}
\end{align}