# 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.

• The term with $2^j=n$ is around $1/e$, so you might still be on course. Jul 21 at 14:41
• @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 ?? Jul 21 at 15:20
• Me neither. Are you sure it isn't a periodic function of $\ln_2 n$? Jul 21 at 15:22
• Where does this problem come from? Jul 21 at 15:27
• @Empy2 as in a Fourier series or something?? Though I am pretty sure the last sum is convergent thought the ratio test. 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 $$O(\exp(-2^{30}))$$.