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 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.
 A: Here's an argument that pushes the lower bound for $S(n)$ closer to a factor of $2$ times
$\log_2 n.$ More precisely, we show
$$ S(n) \ge \left( 2 - \frac{1}{e} \right) \left( \lfloor \log_2 n \rfloor – 1 \right). $$
Let $ a= \lfloor \log_2 n \rfloor $ and write
$ f(n,x) = 1 - \prod_{j=1}^{n-1} ( 1 – jx),$
and so
$$ S(n) = \sum_{k=1}^\infty \left( 1 - \prod_{j=1}^{n-1} \left( 1 - \frac{j}{2^k} \right) \right)$$
$$ = \sum_{k=1}^\infty f(n, \frac{1}{2^k} ) = 
a-1 + \sum_{k=a}^\infty f(n, \frac{1}{2^k} ),$$
since for each $k \le a-1$ $\exists$ $j=2^k$ in $\prod_{j=1}^{n-1} ( 1 - j/2^k ),$ and thus this product is $0.$
So neglecting terms with $k \ge 2a-1,$ we have
$$S(n) \ge a-1 + \sum_{k=a}^{2a-2} f(n, \frac{1}{2^k} ). \qquad (1)$$
By the AM-GM for $x \le 1/(m-1)$ we have for $m \ge 2$ and $2/m(m-1) \le x \le 1/(m-1)$
$$(1-x)(1-2x) \cdots (1-(m-1)x) \le
\left( 1 - \frac{mx}{2} \right)^{m-1} \le
\left( 1 - \frac{1}{m-1} \right)^{m-1} < \frac{1}{e}. $$
Thus for $k=a, a+1,\ldots, 2a-2$ we have
$$ f(n, \frac{1}{2^k} ) \ge 1 - \frac{1}{e}$$
and substituting this into $(1)$ the result follows.
