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.

  • $\begingroup$ The term with $2^j=n$ is around $1/e$, so you might still be on course. $\endgroup$
    – Empy2
    Jul 21 at 14:41
  • 1
    $\begingroup$ @above ah okay wait I think the evaluation of the last sum is not 0 like I erroneously did (since i took the limit inside the summation ) but the thing is how would you compute the last sum first and then apply limits, I don't see any closed form of the last sum ?? $\endgroup$ Jul 21 at 15:20
  • $\begingroup$ Me neither. Are you sure it isn't a periodic function of $\ln_2 n$? $\endgroup$
    – Empy2
    Jul 21 at 15:22
  • $\begingroup$ Where does this problem come from? $\endgroup$
    – VIVID
    Jul 21 at 15:27
  • $\begingroup$ @Empy2 as in a Fourier series or something?? Though I am pretty sure the last sum is convergent thought the ratio test. $\endgroup$ Jul 21 at 15:37

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

enter image description here


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