calculus - How find this sum of $lim_{ntoinfty}sum_{k=1}^{n}left(frac{1}{2^k}sum_{i=0}^{2^{k-1}-1}ln{left(frac{2^k+2+2i}{2^k+1+2i}right)}right)$

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.