Proving $\lim_{n\rightarrow\infty}\sum_{k=1}^n (-1)^{k-1}{n\choose k}\frac{k}{2^k-1}=\frac{1}{\ln2}$ 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.
 A: By elementary binomial coefficient relations one can derive
$$S_n=\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \frac{k}{2^k-1} = n\big( a_n - a_{n-1} \big) $$
where
$$ a_n =\sum_{k=1}^n (-1)^{k-1}\binom{n}{k} \frac{1}{2^k-1}  $$
I've done this because I want to use an asymptotic solution I've presented to the question:
Asymptotics of a recursive sequence
The first terms are simply
$$ a_n \sim \frac{\log{n}}{\log{2}} + \frac{1}{2}+\frac{\gamma}{\log{2}} + \cal{o}(1) $$
Therefore,
$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} n \  \frac{\log{n} - \log{(n-1)} 
}{\log{2}}
= 
\lim_{n \to \infty} n \,  \frac{ -\log{(1-1/n)}}{\log{2}} = \frac{1}{\log{2}}
$$
where a Taylor expansion has been used for the log term.
A: 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 $O(\exp(-2^{30}))$.

