Show that $\sum_{k=1}^{\infty}\frac{2^{-k}}{k}=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}$ without evaluating either sum

Question

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.

Answer 1

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.)

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}