TITLE: How find this sum of $\lim_{n\to\infty}\sum_{k=1}^{n}\left(\frac{1}{2^k}\sum_{i=0}^{2^{k-1}-1}\ln{\left(\frac{2^k+2+2i}{2^k+1+2i}\right)}\right)$
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: Note that
$$
\begin{align}
&\prod_{i=0}^{2^{k-1}-1}\frac{2^k+2+2i}{2^k+1+2i}\\
&=\frac{2^k+2}{2^k+1}\frac{2^k+4}{2^k+3}\frac{2^k+6}{2^k+5}\cdots\frac{2^k+2^k}{2^k+2^k-1}\\
&=\frac{2^k+2}{2^k+1}\color{#A0A0A0}{\frac{2^k+2}{2^k+2}}\frac{2^k+4}{2^k+3}\color{#A0A0A0}{\frac{2^k+4}{2^k+4}}\frac{2^k+6}{2^k+5}\color{#A0A0A0}{\frac{2^k+6}{2^k+6}}\cdots\frac{2^k+2^k}{2^k+2^k-1}\color{#A0A0A0}{\frac{2^k+2^k}{2^k+2^k}}\\
&=\frac{\left(2^{2^{k-1}}2^k!/2^{k-1}!\right)^2}{2^{k+1}!/2^k!}\\
&=\left(\frac1{4^{2^{k-1}}}\left.\binom{2^k}{2^{k-1}}\right)\middle/\left(\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\right)\right.\tag{1}
\end{align}
$$
Furthermore, it is easy to show by induction that
$$
\begin{align}
\prod_{k=1}^n\left(\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\right)^{1/2^{k+1}}
&=\frac1{2^n}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\tag{2}
\end{align}
$$
Therefore, let
$$
f(k)=\frac1{4^{2^k}}\binom{2^{k+1}}{2^k}\tag{3}
$$
Then
$$
\begin{align}
&\sum_{k=1}^n\left(\frac1{2^k}\sum_{i=0}^{2^{k-1}-1}\log\left(\frac{2^k+2+2i}{2^k+1+2i}\right)\right)\\
&=\sum_{k=1}^n\frac1{2^k}\Big(\log(f(k-1))-\log(f(k))\Big)\\
&=\sum_{k=0}^{n-1}\frac1{2^{k+1}}\log(f(k))-\sum_{k=1}^n\frac1{2^k}\log(f(k))\\
&=\frac12\log(f(0))-\frac1{2^n}\log(f(n))-\sum_{k=1}^{n-1}\frac1{2^{k+1}}\log(f(k))\\
&=-\frac12\log(2)-\frac1{2^n}\log(f(n))-\log\left(\frac1{2^n}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\right)\tag{4}
\end{align}
$$
Since Stirling says
$$
\lim_{n\to\infty}\frac1n(n!)^{1/n}=\frac1e\tag{5}
$$
we get
$$
\begin{align}
&\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{2^k}\sum_{i=0}^{2^{k-1}-1}\log\left(\frac{2^k+2+2i}{2^k+1+2i}\right)\right)\\
&=-\frac12\log(2)-\lim_{n\to\infty}\log\left(\frac2{2^{n+1}}\frac{\left(2^{n+1}!\right)^{1/2^{n+1}}}{\sqrt2}\right)\\
&=-\frac12\log(2)-\log\left(\frac{\sqrt2}{e}\right)\\[6pt]
&=1-\log(2)\tag{6}
\end{align}
$$