<s>[INST] Verifying a continued fraction related to $\logφ$. The continued fraction is the following,
$${1+\cfrac{1\cdot 2}{3φ+\cfrac{1\cdot 2}{5+\cfrac{3\cdot 4}{7φ+\cfrac{3\cdot 4}{9+\ddots}}}}}=\frac{2}{3\logφ}\tag{1}$$
Where,
$$φ=\frac{1+\sqrt{5}}{2}$$
Something with I found in an 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)?
 [/INST] 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 $$\sinh^{-1}\frac{1}{\sqrt\varphi}=\log\left(\frac{1}{\sqrt\varphi}+\sqrt{\frac{1}{\varphi}+1}\right)=\log\left[\sqrt\varphi\left(\frac{1}{\varphi}+1\right)\right]=\frac32\log\varphi.$$
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