Sum and Product of Infinite Radicals How to compute $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}$$ & $$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}$$?
I understand that  $$\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots$$
and
$$\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} $$
How to Proceed further?

About
$$ c=\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\ldots}}}} $$
we may notice that for any $n\geq 1$ the functions $f_n(x)=\sqrt{n+x}$ are contractions of $[\varphi,+\infty)$ with Lipschitz constant $\leq\frac{1}{2\sqrt{n+1}}$. Additionally, for any $n\geq 2$ we have
$$ \sqrt{n+\sqrt{n+1+\sqrt{n+2+\sqrt{n+3+\ldots}}}}<\sqrt{n+\sqrt{n^2+\sqrt{n^4+\sqrt{n^8+\ldots}}}}=\varphi\sqrt{n}$$
so
$$ \left|c-\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4}}}}\right|\leq\frac{\varphi}{2^4\sqrt{4!}} $$
and, in general, the "truncations" of the nested radical defining $c$ converge extremely fast to $c$:
$$ \left|c-\sqrt{1+\sqrt{2+\sqrt{\ldots+\sqrt{n}}}}\right|\leq\frac{\varphi}{2^n \sqrt{n!}}. $$
By considering $n=10$ we already get five correct figures: $c\approx 1.75793$.
