Verifying a continued fraction related to $\log\varphi$.

Question

The continued fraction is the following, $${1+\cfrac{1\cdot 2}{3\varphi+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7\varphi+\cfrac{3\cdot 4}{9+\ddots}}}}}=\frac{2}{3\log\varphi}\tag{1}$$ where $$\varphi=\frac{1+\sqrt{5}}{2},$$ something with I found in an aside [backside] of my previous notes where I kept my recreational math works. So can the closed form be verified in any sort (most preferably by using already established identities)?

Answer 1

This can be deduced from Gauss's continued fraction for $_2F_1$, written as $$\frac{_2F_1(a+1,b;c+1;z)}{_2F_1(a,b;c;z)}=\cfrac{c}{c+\cfrac{(a-c)bz}{c+1+\cfrac{(b-c-1)(a+1)z}{c+2+\cfrac{(a-c-1)(b+1)z}{c+3+\cfrac{(b-c-2)(a+2)z}{c+4+\ddots}}}}}.$$ With $a=b=1$ and $c=3/2$, computing the values of $_2F_1$, we obtain $$1+\cfrac{1\cdot 2}{3/z+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7/z+\cfrac{3\cdot 4}{9+\ddots}}}}=1+\frac{2z}{3}\frac{_2F_1(2,1;5/2;-z)}{_2F_1(1,1;3/2;-z)}=\frac{\sqrt{z(1+z)}}{\sinh^{-1}\sqrt{z}}.$$ Now we put $z=1/\varphi=\varphi-1$ and note that $$\sinh^{-1}\frac{1}{\sqrt\varphi}=\log\left(\frac{1}{\sqrt\varphi}+\sqrt{\frac{1}{\varphi}+1}\right)=\log\left[\sqrt\varphi\left(\frac{1}{\varphi}+1\right)\right]=\frac32\log\varphi.$$