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?
 A: $$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\cdots}}}}=C$$
Where $C \approx 1.7579$ (OEIS A072449) is the nested radical constant.  No closed form of this constant is known.

$$\sqrt{1\sqrt{2\sqrt{3\sqrt{4\cdots}}}}=\sigma$$
where $\sigma \approx 1.6616$ (OEIS A112302) is Somos's Quadratic Recurrence Constant.
$\sigma$ has an alternate form (I hesitate to call it a "closed form") of 
$$\sigma =
\exp\left[-2^n \frac{\partial \operatorname{Li_n\left(\frac{1}{2}\right)}}{\partial n}\bigg|_{n=0}+\frac{1}{2} \frac{\partial \Phi\left(\frac{1}{2},s,n+1\right)}{\partial s}\bigg|_{s=0}\right]$$
where $\operatorname{Li}_n(z) = \sum^\infty_{n=1} \frac{z^k}{k^n}$ is the polylogarithm and $\Phi(z,s,a)=\sum_{k=0}^\infty \frac{z^k}{(a+k)^s}$ is the Lerch transcendent, and 
$$\sigma = \exp\left[\int_0^1 \frac{1-x}{(x-2)\log x}\, \mathrm{d}x\right]$$
A: For the first:
Well, if it converges, we may compute its logarithm.
Thus, $\log x = \sum_{k=1}^\infty \frac{\log k}{2^k}$
Now, looking at this, it is evident that this actually DO converge, (the numerators are eventually dominated by $1.5^k$ so
$\log x < K + \sum_{k=1}^\infty \frac{1.5^k}{2^k}$
for some finite constant $K.$ Now, this sum is geometric and converges).
Now, computing the VALUE of this is trickier, and it is most probably not a rational number or some "nice" expression. In fact, $x$ converges to about 1.66169 (Mathematica), and is the exponential of some expression involving a Polylogarithm.
EDIT:
Mixing recursion and different arithmetic operations usually result in constants that have no nice expressions. For example, $1/F_1 + 1/F_2 + 1/F_4+\dots$ converges to a number which is unknown to be rational or not, irrational $F_i$ is the i'th Fibonacci number, http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant
A: 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: $c\approx 1.75793$.
A: Let $A=(1^{1/2})(2^{1/4})(3^{1/8})...\implies \log A= \sum_{i=1}^{\infty}\frac{\log i}{2^i}$ and this sum converges to $\approx 0.507834$
http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427em8adj6o2qa
$\implies (1^{1/2})(2^{1/4})(3^{1/8})...$ converges to $\approx e^{0.507834}$. 
