This is a follow up of Upper bound binomial sum
I was working on the problem in the above thread and noticed an interesting thing. I wanted to try and improve the bound Derek had (which was a quadratic in $n$).
If we reformulate the problem as Derek has (since for this we need $2^k \geq n$, so we forget the original problem and think the problem given is as follows)
i.e.
Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$.
We see that
$S(1) = 0$
$S(2) = 1$
$S(3) = \frac{7}{3} \approx 2.3333$
$S(4) = \frac{67}{21} \approx 3.1904$
$S(5) = \frac{407}{105} \approx 3.8762$
$S(6) = \frac{4789}{1085} \approx 4.4138$
$S(7) = \frac{5289}{1085} \approx 4.8747$
$S(8) = \frac{726093}{137795} \approx 5.2694$
$S(9) = \frac{118399669}{21082635} \approx 5.61598$
$S(10) = \frac{9120486643}{1539032355} \approx 5.92612$
$S(11) = \frac{105065594573}{16929355905} \approx 6.20612$
$S(12) = \frac{31986101239583}{4950627362505} \approx 6.4610$
We see that the growth is slow as $n$ increases. My question is if this converges to some limit or it diverges. I have been working on it for the past couple of hours but am unable to come to a conclusion.
So the question is what is $\displaystyle \lim_{n \rightarrow \infty} S(n)$?
If it converges, can we find the limit?
or
If it diverges, how fast does it diverge?
$\textbf{EDIT:}$
So Mike has proved that $S(n)$ in fact diverges.
The conjecture is now that $\lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$ where $\alpha \approx 0.667$.
Look at Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$ - Part II for further discussions.