Infinite product involving powers of 2 I have the following infinite product:
$2/1 * 3/2 * 5/4 * 9/8 * 17/16 * 33/32 * 65/64...$
What does it converge to?
I can take its $\ln()$ to get
$\ln(2) + \ln(3/2) + \ln(5/4)....$
Which using the quotient difference rule resolves to
$S(n) = \ln(2^n + 1) - n\ln(2)$ for $n = 0$ to $\infty$
At this point I don't know how to evaluate this: so I took its derivative to see if I could recognize anything:
$S'(n) = 2^n * ln(2) / (2^n + 1) - ln(2)$
Which really looks like it converges to 0... 
Integrating $S(n)$ hasn't been much to avail unless its because I lack knowledge.
What's this come out to?

Drop the first term for a moment.
$$
\frac{3}{2}\frac{5}{4}\cdots\frac{2^k+1}{2^k}\cdots = \prod_{k=1}^{\infty} (1 + \frac{1}{2^k}) 
$$
When the product is expanded, what is the coefficient of $\frac{1}{2^n}$? A term comes from any number of distinct factors of the form $\frac1{2^k}$.  And so, each partition
$$
n = k_1 + k_2 + \cdots + k_m
$$
provides a term in the $\frac{1}{2^n}$.  So, the product is:
$$
\sum_{n=0}^{\infty} \frac{q(n)}{2^n}
$$
where $q(n)$ is the number of partitions of $n$ into distinct integers. I don't think this has a closed form.
