This nested radical is one that I keep unsuccessfully working on solving. See here $$G(0)=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\cdots}}}}$$ In recurrence form, this can be written as $$G(x)^2=2^{x-1}+G(x+1)$$ and then evaluated at $x=1$
Ramanujan showed that for some nested radical of the form $$F(x)^2=ax+(a+n)^2+xF(x+n)$$ then $$F(x)=a+x+n$$
To get $G(x)$ into that form, we can say $$G(x)^2=\frac{2^{x-1}}{j(x-1)}+xG(x+1)$$ where $$j(x)=\prod_{i=1}^x (x-i+1)^{({2^i})}$$ which can be defined as $$j(x)=(xj(x-1))^2$$ with $$j(0)=1$$ Setting $n=1$, I thought that $F(x)=G(x)$ when $$ax+(a+1)^2=\frac{2^{x-1}}{j(x-1)}$$ Naturally I want to set $x=1$ because I want to find $G(1)$, but then we'll have $a=0$. Putting everything together we'd get $F(1)=0+1+1\ne{G(1)}$.
Trying different values of $x$ and $a$ and then correcting for the required iterations doesn't work either. What's wrong?
