Infinitely nested radical problem?

Question

I became interested in this nested radical from another question and thought I would have a go at trying to come up with a formula for it. It is $$G(0)=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\cdots}}}}.$$ This can be written as a recurrence relation when evaluated at $x=0$: $$G(x)^2=2^x+G(x+1).$$ Due to Ramanujan, we know that for some function $F(x)$ such that $$F(x)^2=ax+(a+n)^2+xF(x+n)$$ we have $$F(x)=a+n+x.$$ We can put $G(x)$ and $F(x)$ into equatable forms by making $x=1, n=1, a=\frac{\sqrt{13}-3}{2}.$ Summing these and correcting for $G(0)$ we should have $$G(0)=\sqrt{1+\frac{\sqrt{13}-3}{2}+1+1}=\sqrt{\frac{\sqrt{13}+3}{2}}.$$ this gives $1.817354021023971.$ However the correct value for $G(0)$ is around $1.78316580926410.$ What is the error in my reasoning?

Answer 1

Your mistake was assuming that $F(x)$ and $G(x)$ are the same for any $x$ simply because they are the same for $x=1$:

$$G(1)=\sqrt{2^1+F(1+1)}=\sqrt{2+F(2)} \\ F(1)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(1)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(1)F(1+1)}=\sqrt{2+F(2)}$$

But look at $x=2$: $$G(2)=\sqrt{2^2+F(2+1)}=\sqrt{4+F(3)} \\ F(2)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(2)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(2)F(2+1)} \approx \sqrt{2.3028+2F(3)}$$

So if you replace $G(x+1)$ with $F(x+1)$ in your definition of $G$ (as you did to solve), then $G(0)$ actually looks like this when expanded: $$G(0)=\sqrt{\frac{\sqrt{13}+3}{2}} \approx\sqrt{1+\sqrt{2+\sqrt{2.3028+2\sqrt{2.6055+\dots}}}}$$