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?

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    $\begingroup$ Is is trivial that $\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+\ldots}}}}}$ converges to a finite value? It is not to me, at least. $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 18:05
  • $\begingroup$ That is trivial I think. The finite value is what I want to find. Why is it not trivial to you? $\endgroup$ – tyobrien Jan 24 '17 at 18:08
  • $\begingroup$ Probable duplicate of math.stackexchange.com/q/1274352/269624 $\endgroup$ – Yuriy S Jan 24 '17 at 18:09
  • $\begingroup$ @TyO'Brien: well, if it is trivial: why $\sqrt{1+\sqrt{2+\sqrt{4+\ldots}}}$ converges to a finite value? $\endgroup$ – Jack D'Aurizio Jan 24 '17 at 18:18
  • $\begingroup$ As I've already hinted at about a year and a half ago, in the comment section of the post linked to in the very body of the question itself, even OEIS has no earthly idea about a possible closed form for this infinitely nested expression. $\endgroup$ – Lucian Jan 24 '17 at 18:35

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