Find $\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$ 
Find the value of $$\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots}}}}$$

I know how to solve when all surds are of the same order, but what if they are different?
Technically, (as some users wanted to know exactly what is to be found), find:
$$\lim_{n\to\infty}\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\cdots+\sqrt[n]{4}}}}} $$
 A: It's a bit longer comment so I'll write here (but it's not a solution so I'm deleting when a real answer appears).
I'm thinking along the lines of the sequence
$$f_1=x+4$$
$$f_2=(f_1-4)^2$$
$$f_n=(f_{n-1}-4)^n$$
This is the sequence of radicands, which is divergent for every $x$ except for the one that solves our question. So we're looking for $x$ for which this sequence is monotonically convergent down to $5$. Any minor deviation, and it starts diverging. So we're looking for an unstable critical point of the sequence above.
$(f_n)$ is a sequence of polynomials in $x$:
$$x+4,x^2,x^6-12x^4+48x^2-64,x^{24}-48x^{22}+\cdots$$
The solution is $x$, such as $\lim_{n\to\infty}f_n=5$ exists, but we could restate it as a convergence criterion $\lim_{n\to\infty}\frac{f_{n+1}}{f_n}= 1$. I was hoping that in the limit, only a subset of polynomial coefficients with known asymptotics matter, but here I got stuck, because it appears all of them are important.
Another way of looking at this is to find the largest root of $f_n-5=0$ in the limit $n\to\infty$.
I figured out the first order asymptotics for $f_n$ at the correct $x$:
$$f_n\asymp 5 + \frac{\ln 5}{n}$$
Any thoughts on this approach?
A: Put $$y=\sqrt{4+\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...}}}}\qquad (1)$$ stopping successively at the $n$-th root we have a sequence  strictly increasing and bounded  so the limit $y$ is well defined.
From $(1)$ we can easily get a sequence $\{P_n\}$ of polynomials such that its largest real roots $$\alpha_n=\sqrt{4+(4+(4+(4+….(4+\sqrt[n]4)^{1/n}….)^{1/5})^{1/4})^{1/3}}$$
form a sequence $\{\alpha_n\}$ converging to $y$. In fact we have
$$\begin{cases}P_2(x)=x^2-4\\P_3(x)=(x^2-4)^3-4\\P_4(x)=( (x^2-4)^3-4)^4-4\\.....\\.....\\P_n(x)=(P_{n-1}(x))^n-4\end{cases}$$
We can see the polynomial $P_n$ has degree $n!$ and $P_n(\alpha_{n-1})=-4$. Since $P_n(x)$  grows very quickly, this indicates that $\{\alpha_n\}$  rapidly converges to the limit $y$ because the arc of the graphic of $P_n(x)$ between the points $(\alpha_{n-1},-4)$ and $(\alpha_n,0)$ is almost a vertical line so $\alpha_{n-1}\approx \alpha_n$. This argument is enhanced by the equality 
$$y=2\sqrt{1+\left(\frac{4}{4^3}+\left(\frac{4}{4^{12}}+\left(\frac{4}{4^{60}}+….\left(\frac{4}{4^n}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ hence the factor of $2$ is equal to $$\sqrt{1+\left(\frac{1}{2^{6-2}}+\left(\frac{1}{2^{24-2}}+\left(\frac{1}{2^{120-2}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$ or
$$\sqrt{1+\left(\frac{1}{2^{4}}+\left(\frac{1}{2^{22}}+\left(\frac{1}{2^{118}}+….\left(\frac{1}{2^{n!-2}}+\sqrt[n]4\right)^{1/n}….\right)^{1/5}\right)^{1/4}\right)^{1/3}}$$  This factor of $2$ grows very slowly obviously.
It seems to me there is not a closed form for $y$ (It could be an asymptotic one perhaps). Consequently I give here an (justified) approximation taking the above referred real root of, say, $P_6$; we have$$y\approx\sqrt{4+\left(4+\left(4+\left(4+\left(4+\sqrt[6]4\right)^{1/6}\right)^{1/5}\right)^{1/4}\right)^{1/3}}\approx \color{red}{2.40161550315}$$  
