TITLE: Proving $\lim_{n\rightarrow\infty}\sum_{k=1}^n (-1)^{k-1}{n\choose k}\frac{k}{2^k-1}=\frac{1}{\ln2}$
QUESTION [12 upvotes]: Can someone help with this sum?

Prove $$S=\lim_{n\rightarrow\infty}\sum_{k=1}^n (-1)^{k-1}{n\choose k}\frac{k}{2^k-1}=\frac{1}{\ln2}$$

I have tried to break down the $\frac{k}{2^k-1}=\sum_{j=1}^\infty \frac{k}{2^{kj}}$ and tried writing it as $$\lim_{n\rightarrow\infty}\sum_{j=1}^\infty\sum_{k=1}^n (-1)^{k-1}{n\choose k}k\left(\frac{1}{2^{j}}\right)^k,$$then using the binomial summation as such, note $\left(x=\frac{1}{2^j}\right)$
$$\lim_{n\rightarrow\infty}\sum_{j=1}^\infty\frac{1}{2^j}\frac{\mathrm{d}}{\mathrm{d}x}\left(\sum_{k=1}^n (-1)^{k-1}{n\choose k}x^{k}\right)=\lim_{n\rightarrow\infty}\sum_{j=1}^\infty\frac{n(1-x)^{n-1}}{2^j}.$$ Giving us the final sum to be $$\lim_{n\rightarrow\infty}\sum_{j=1}^\infty\frac{n\left(1-\frac{1}{2^j}\right)^{n-1}}{2^j},$$ the limit of which seems to be $0$. Which is not what we want, can someone tell me where my mistakes lie, and also point towards the solution or present it themselves. Thanks in Advance.

REPLY [2 votes]: The sum seems to vary at the fifth decimal place.  I found a similar result in What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.
As I said above, in comments, the sum seems to be
$$\sum_{j=1}^{\infty}\frac n{2^j}\exp\left(-\frac n{2^j}\right)$$
When $n$ is large, the dominant terms are when $2^j$ is close to $n$, then the terms in both directions (both small $j$ and large $j$) approach zero.  So the sum becomes
$$\sum_{k=-\infty}^{\infty} 2^{k+\log_2 n}\exp\left(-2^{k+\log_2 n}\right)$$
and the sum is the function of the fractional part of $\log_2 n$.
The graph below shows this sum varies at the fifth decimal place - the amplitude is around $0.00001426$.  This is not an artifact.  I took $k+\log_2n$ ranged between $-30$ and $30$. The missing terms for $k+\ln n\lt-30$ are $O(2^{-30})$, and for $k+\ln n\gt30$ are $O(\exp(-2^{30}))$.