On wiki page Naturel logarithhm of 2, other representations section it lists $$\ln\,2=\cfrac 1{1+\cfrac 1{2+\cfrac 1{3+\cfrac 2{2+\cfrac 2{5+\cfrac 3{2+\cfrac 3{7+\cfrac 4{2+\cfrac 4{9+\cdots}}}}}}}}}$$
How to derive this? I checked Euler's continued fraction formula page but still couldn't figure out how the $a_i$ and $b_i$ are derived.
A math stackexchange answer mentioned $$\;\left[\matrix {n_k\\d_k}\right]=\left[\matrix {n_{k-1}\;n_{k-2}\\d_{k-1}\;d_{k-2}}\right]\left[\matrix {b_k\\a_k}\right],\quad\left[\matrix {n_1\\d_1}\right]=\left[\matrix {b_0\,b_1+a_1\\b_1}\right]=\left[\matrix {1\\1}\right],\;\left[\matrix {n_0\\d_0}\right]=\left[\matrix {b_0\\1}\right]=\left[\matrix {0\\1}\right]$$ but didn't explain where this is from.
The given continued fraction, rewritten as $$L=\cfrac{1}{1+\cfrac{1^2}{2+\cfrac{1^2}{3+\cfrac{2^2}{4+\cfrac{2^2}{5+\cfrac{3^2}{6+\cfrac{3^2}{7+\ddots}}}}}}},$$ is a special case of Gauss's continued fraction for $_2F_1$ (taken at $a=0$, $b=c=1$, $z=-1$): $$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}.$$
Now $_2F_1(0,1;1;z)=1$ and $_2F_1(1,1;2;z)=-\ln(1-z)/z$ give the expected $L=\ln 2$.