I did not found it on the forum so :
$$S=\sqrt{1+\frac{1}{2}\sqrt{1+\frac{1}{2^2}\sqrt{1+\frac{1}{2^3}\sqrt{1+\frac{1}{2^4}\sqrt{\cdots}}}}}=1.25$$
I try to denested the radical but I'm really stuck to get wich follows .
Maybe a generalization could be give the result.
Furhtermore I have a question is 2 the only number such that :
$$\sqrt{1+\frac{1}{k}\sqrt{1+\frac{1}{k^2}\sqrt{1+\frac{1}{k^3}\sqrt{1+\frac{1}{k^4}\sqrt{\cdots}}}}}$$
is rational ?
Where $k\geq 2$ is a natural number
Thanks a lot for your answer and your time .
Best Answer
Consider the unique power series in $\,t\,$ that satisfies $$ f(t,q) = \sqrt{1 + tf(t\,q,q)}. \tag{1} $$ We can express it as a continued square root $$ f(t,q) = \sqrt{1 + t\sqrt{1 + t\,q\sqrt{1 + t\,q^2 \cdots}}}. \tag{2} $$ The first few terms are $$ f(t,q) \!=\! 1 \!+\! (1/2)\,t^2 \!+\! (-1/8\!+\!1/4q)\,t^3 \!+\! \\ (1/16q\!-\!1/8q\!-\!1/16q^2\!+\!1/8q^3)\,t^4 \!+ \cdots .\tag{3} $$ A special case closed form formula is $$ f(1/2^n,1/2)=1+2^{-(n+1)}, \tag{4} $$ proven by induction using equation $(1)$ and the identity $$(1 + r/2) = \sqrt{1 + r(1 + r/4)}\;\; \text{ where } \;\; r = 2^{-n}. \tag{5} $$ The question asked about $\,f(\frac12,\frac12) = \frac54\,$ using equation $(4)$ with $\,n=1.$
NOTE: In equation $(2)$ the continued square root is the limit of a sequence $$ x,\; \sqrt{1\!+\!t\,x},\; \sqrt{1\!+\!t\sqrt{1\!+\!t\,q\,x}},\; \sqrt{1\!+\!t\sqrt{1\!+\!t\,q\sqrt{1\!+\!t\,q^2x}}},\;\cdots $$ which converges at least for non-negative $\,t,q,x\,$ which needs some kind of proof in general.