Show that $\sum_{k=1}^{\infty}\frac{2^{-k}}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}$ without evaluating either sum 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. 
 A: Use the Euler transform on $\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}$:
$$\begin{split}\sum_{k=0}^{\infty}\frac{(-1)^k}{k+1}&=\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{k+1}}\left.\Delta^k\frac{1}{m+1}\right\rvert_{m=0}\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{k+1}}\left.\frac{(-1)^k}{\binom{m+k+1}{k}}\right\rvert_{m=0}\\
&=\sum_{k=0}^{\infty}\frac{(-1)^k}{2^{k+1}}\frac{(-1)^k}{k+1}\\
&=\sum_{k=0}^{\infty}\frac{1}{2^{k+1}(k+1)}\text{.}\end{split}$$
(The key equality
$$\Delta^k\frac{1}{m+1}=\frac{(-1)^k}{\binom{m+k+1}{k}}$$
can be shown inductively.)
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
A: Not an answer, only some thoughs about it, too long for a comment.

$\begin{align}\sum\limits_{n=1}^{2N}\dfrac{(-1)^{n-1}}n &=\sum\limits_{n=1}^N\left(\dfrac 1{2n-1}-\dfrac 1{2n}\right) =\sum\limits_{n=1}^N\left(\dfrac 1{2n}+\dfrac 1{2n-1}-\dfrac 1{n}\right)\\ &=\sum\limits_{n=1}^{2N} \dfrac 1n-\sum\limits_{n=1}^N\dfrac 1n =\sum\limits_{n=N+1}^{2N}\dfrac 1n=\sum\limits_{n=1}^{N}\dfrac 1{n+N}\\ &=\dfrac 1N\sum\limits_{n=1}^{N}\dfrac 1{1+\frac nN}\xrightarrow{\text{Riemann sum}}\int_0^1\dfrac{dt}{1+t}=\ln(2)\end{align}$
Now with your problem you get $\displaystyle\sum\limits_{k=1}^{\infty}\dfrac 1{2^kk}=\int_0^1\dfrac{\mathop{du}}{2-u}$
Both integrals are equal by the change of variable $x=1-u$
We can also easily report in the Riemann sum to get $\sum\limits_{n=1}^N\dfrac 1{2N-n}=\sum\limits_{n=N}^{2N-1}\dfrac 1n$ and get back to the alternate series.
But I do not see a direct way to go from $\sum\limits_{n=1}^N\dfrac 1{2N-n}$ to $\sum\limits_{k=1}^{??}\dfrac 1{2^kk}$, I suppose that since the convergence is greatly accelerated, we have to sum a lot of $(-1)^k$ terms for each single $2^{-k}$ one...
I too wonder if we can prove this directly without passing via the integrals.
