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.

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.

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 $O(\exp(-2^{30}))$.