TITLE: 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
QUESTION [7 upvotes]: This is a follow up of Limit of $S(n) = \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$
More details can be found in the above thread.
Let $S(n) = \displaystyle \sum_{k=1}^{\infty} \left(1 - \prod_{j=1}^{n-1}\left(1-\frac{j}{2^k}\right)\right)$
Mike has proved that $S(n)$ in fact diverges at-least faster than $\log_2(\lfloor n-1 \rfloor)$.
Now based on what Mike has worked this conjectures arises:
$\displaystyle \lim_{n \rightarrow \infty} (2 \log_{2}(n) - S(n)) = \alpha$.
Also, can $\alpha$ be expressed in terms of other familiar constants. $\frac{\pi \gamma}{e}$ seems to be a close guess.
The numerical evidence seem to suggest they are true.  For example, we have the following graph of $2 \log_2 n - S(n)$ for $n \leq 300$.  

(More numerical evidence: The value of $2 \log_2 n - S(n)$, is, for $n = 1000$, $2000$, and $3000$, respectively, $0.667734$, $0.667494$, and $0.667413$.)
An alternative expression for $S(n)$ was worked out by Moron in the previously-mentioned question:
$$S(n) = - \sum_{k=1}^{n-1} \frac{s(n,k)}{2^{n-k}-1},$$ 
where $s(n,k)$ is a Stirling number of the first kind.

REPLY [9 votes]: Here's a proof.  The value of $\alpha$ is the value of the infinite sum $$\sum_{m= -\infty}^{\infty} (e^{-2^{-m-1}} - [m \geq 1]),$$  where $[m \geq 1]$ is 1 if $m \geq 1$ and 0 otherwise.  Mathematica gives this value (to 6 decimal places) as $0.667253$.
The full argument in all its rigor is too long to post on this site, so I'm only going to give an extended outline.  There are a couple of strange claims in here, but bear with me.  

Part 1
From my previous post we know that $1 - \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) = 1$ when $k \leq \lfloor \log_2 (n-1) \rfloor$.  
Let $p = 2 \log_2 n - \lfloor 2 \log_2 n \rfloor$.    Thus  $2 \log_2 n - S(n)$ is  $$2 \log_2 n - \lfloor 2 \log_2 n \rfloor + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{\lfloor 2 \log_2 n \rfloor} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= \lfloor 2 \log_2 n \rfloor + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right)$$
$$= p + \sum_{k= \lfloor \log_2 (n-1) \rfloor +1}^{2 \log_2 n - p} \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) + \sum_{k= 2 \log_2 n - p + 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right) - 1\right) .$$
Now, the expression $$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^k}\right)$$ is very close to 0 when $k < 2 \log_2 n$ and very close to 1 when $k > 2 \log_2 n$.  So most of the contribution to $2 \log_2 n - S(n)$ occurs when $k$ is close to $2 \log_2 n$.  The next step, then, is to reindex with $m = k - \lfloor 2\log_2 n \rfloor$.  Now we basically have 
$$2 \log_2 n - S(n) = p + \sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) + \sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right).$$

Part 2
Next, we need a good approximation to $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$.  It turns out that $e^{-2^{-m+p-1}}$ is an excellent approximation (which surprises me some - despite the fact that I have verified it numerically - as it is independent of $n$).  To see this, rewrite $\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right)$ as $$\exp \left( \sum_{j=1}^{n-1} \ln\left(1 - \frac{j}{2^{m-p} n^2}\right)\right)$$ and then expand the $\log$ expression with the Maclaurin series for $\ln (1+x)$.  The first term in the expansion dominates when $m$ is positive or constant, and we get
$$\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \exp \left(\sum_{j=1}^{n-1} \left(- \frac{j}{2^{m-p} n^2} \right) + O\left(\frac{j^2}{4^{m-p} n^4}\right) \right)$$
$$=\exp \left( - \frac{1}{2^{m-p+1}} + O\left(\frac{1}{2^m n}\right) \right) = \exp \left( - \frac{1}{2^{m-p+1}}\right) + O\left(\frac{1}{2^m n}\right)$$
Thus $$\sum_{m= 1}^{\infty} \left(\prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) - 1\right) = \sum_{m= 1}^{\infty} \left(\exp \left( - \frac{1}{2^{m-p+1}} \right) - 1\right) + O\left(\frac{1}{n}\right).$$
When $m$ is negative and not constant, things are a little trickier, as the higher-order Maclaurin series terms make $\exp \left( - \frac{1}{2^{m-p+1} } \right)$ not a good relative approximation to the product.  However, all the terms in the Maclaurin series are negative, so truncating after the first term does yield an upper bound.  In addition, $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ goes to zero extremely fast as $m \to -\infty$.  (For example, if $m = - \log_2 (\log n)$ (which heads to $-\infty$ very slowly), we still have $\exp \left( - \frac{1}{2^{m-p+1}} \right) = \frac{1}{n^{2^{p-1}}}$.)  Thus $\exp \left( - \frac{1}{2^{m-p+1}} \right)$ is still an excellent absolute approximation for the product when $m$ is negative.  Since there are only $\log_2 n$ negative terms in the sum we are considering, we have 
$$\sum_{m= - \log_2 n}^0 \prod_{j=1}^{n-1} \left(1 - \frac{j}{2^{m-p} n^2}\right) = \sum_{m = - \log_2 n}^0 \exp \left( - \frac{1}{2^{m-p+1}} \right) + E(n),$$
where $E(n) \to 0$ as $n \to \infty$.

Part 3
Now, let $$f(p) = \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1).$$
The function $f$ is linear in $p$!  (Despite having verified this numerically and despite the argument below, this still seems weird to me!)  The slope is $-1$.  To see this, differentiate to get $$f'(p) = - \ln 2 \sum_{m= - \infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}}.$$ 
Now, apply the Euler-Maclaurin summation formula.  Because we have the product of two exponentials, $f^{(k)}(p)$ will have the expression $2^{-m+p-1} e^{-2^{-m+p-1}}$ in every term.  As $m \to \infty$, $2^{-m+p-1} \to 0$ and $e^{-2^{-m+p-1}} \to 1$.  As $m \to -\infty$, $2^{-m+p-1} \to \infty$ and $e^{-2^{-m+p-1}} \to 0$, but the latter expression dominates.  Thus  $f^{(k)}(p) \to 0$ as $m \to \infty$ and as $m \to -\infty$.  Thus the Euler-Maclaurin formula says that
$$f'(p) = - \ln 2 \int_{-\infty}^{\infty} 2^{-m+p-1} e^{-2^{-m+p-1}} dp = \left. e^{-2^{-m+p-1}}\right|_{-\infty}^{\infty} = -1.$$ 
Therefore, $$\lim_{n \to \infty} \left(2 \log_2 n - S(n)\right) = p + \sum_{m= - \infty}^0  e^{-2^{-m+p-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m+p-1}} - 1) $$
$$= p - p + \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1) = \sum_{m= - \infty}^0  e^{-2^{-m-1}} + \sum_{m= 1}^{\infty} (e^{-2^{-m-1}} - 1).$$
Again, this value is approximately $0.667253$.