How find this sum of $\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{2^k}\sum_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}\right)$ Find the sum  of the limit
$$\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{2^k}\sum_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}\right)$$
My try: since
$$\sum_{i=0}^{2^{k-1}-1}\ln{\left(\dfrac{2^k+2+2i}{2^k+1+2i}\right)}=\sum_{i=0}^{2^{k-1}-1}\left(\ln{(2^k+2+2i)}-\ln{(2^k+1+2i)}\right)$$
My friend tells me this sum has an analytical solution. But
I can't find it. Thank you.
 A: Use
$$\ln\left(\frac{2^k+2+2i}{2^k+1+2i}\right)=\ln\left(1+\frac{1}{2^k+1+2i}\right)$$
and $\ln(1+u)=u-\tfrac12 u^2+\tfrac13u^3\mp...$

Probably more tricky than that. Write down the sums up to $n=4$ in your transformation of the logarithms and consider that for example 
$\ln(14)=\ln(2)+\ln(7)$ and $\ln(28)=2\ln(2)+\ln(7)$. 
I'd expect that the logarithms of the odd numbers cancel in the long run, however I do not see what the buildup of the $\ln(2)$ terms leads to.

Expanded: Define $$A_k=\sum_{i=0}^{2^{k-1}-1}\ln(2^k+2i+2)$$ for the sum of logs of even numbers between $2^k+1$ and $2^{k+1}$ and $$B_k=\sum_{i=0}^{2^{k-1}-1}\ln(2^k+2i+1)$$ for the sum of logs of odd numbers in the same segment. Then the task is to determine the limit for
$$\sum_{k=1}^n \frac{A_k-B_k}{2^k}$$
But we also have the recursion that the even numbers can be divided by $2$ and thus mapped to the previous interval, $$A_{k+1}=2^k\ln(2)+A_k+B_k,$$ so the big sum reduces to
$$\sum_{k=1}^n \frac{2^k\ln2+2A_k-A_{k+1}}{2^k}=n\ln(2)+A_1-\frac{A_{n+1}}{2^n}$$
and for the last term we can finally reach back to the Riemann sum techniques.

I may have overlooked something, ... according to my quick'n'dirty calculation the result could be just simply $$1-\ln(2).$$
A: studying the terms of the series more carefully (after my initial crass misreading!) it occurred to me that we are dealing here with an assertion about geometric means.
let $$s_n=\sum_{k=1}^{n}\frac{\sum\limits_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}}{2^k}
$$ 
and define $g_n$ to be the geometric mean of the $2^n$ consecutive integers $2^n+1, 2^n+2,\dots,2^{n+1}$. then we have (something like):
$$ e^{s_n} = \frac{2^n}{g_n}$$
to find a more precise value we may write:
$$ g_n^{2^n}= \frac{2^{n+1}!}{2^n!}
$$ to which we may apply Stirling's approximation for the gamma function.  I am happy to leave the more intricate details of the calculation to others less prone to silly mistakes. however a quick back-of-the envelope calculation suggests that:
$$\lim_{n\to\infty} s_n = \frac{e}4
$$
note to self. 1)forgot to take logs at the end (pretend I ran out of envelope) 2) forgot 1st numerator doesn't cancel! 
