Value of $y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$ I was given this problem on series by a friend. 
If 
$$y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$$
then how to solve such problem.
I don’t want the full answer, rather, insights, mathematical facts, theorems, and relationships that would help me solve it on my own.
My efforts: I thought that the whole thing inside the square bracket must be a perfect square so we have [$4~+$ something] should be a positive perfect square but that would be like finding a trivial solution by trial and error method so I don't know how to solve it.
I also tried by squaring and checking like this 
$$y^2 - 4=\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}$$
so we get two factors $y-2$ and $y+2$, but still it was like same trial and error method of finding factors . So can any one help.
Thanks in advance.
Edit:  The only reasonable interpretation is the recurrence $y_n=\sqrt[n]{4+y_{n+1}}$ (Thanks @par)
 A: Too long for a comment, so
$\begin{array}\\
y
&=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{\sqrt[4]{4}}\sqrt[5]{4}\sqrt[5]{1+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}}\sqrt[3]{1+\frac1{4^{1/12}}\sqrt[4]{1+\frac1{4^{1/20}}\sqrt[5]{1+...............}}}}\\
\end{array}
$
It looks like
there is a pattern of
$\dfrac1{4^{1/n-1/(n+1)}}
=\dfrac1{4^{1/(n(n+1))}}
$
which might make it easier
to get a more solvable recurrence.
And, of course,
$4$ can be replaced
by any value,
probably preferably a square.
A: This is only a hint, because the following argumentation is a rough calculation. 
$y^n_n=4+y_{n+1}$ 
If we assume a value $>1$ for $y^n_n$ we can set $y_n\approx 1+\frac{a}{n}$ for large $n$ so that we get 
$(1+\frac{a}{n})^n\approx 4+(1+\frac{a}{n+1})$ which means for $n\to\infty $ the equation $e^a=5$ or simply 
$a=\ln 5\approx 1.6094379...$ . 
It follows (for small $n$, here $n:=2$) $\enspace y:=y_2\approx \sqrt{4+(1+\frac{\ln 5}{2})}\approx 2,409298...$ .
I think this value is a good upper bound for $y$ . 
A lower bound should be $\sqrt{4+(1+\frac{\ln 5}{3})}\approx 2,35297...$ .  
A: The question is pretty close to this one, just replace $\,2\,$ by $\,4\,$ in:
Evaluating the limit of $\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$ when $n\to\infty$
There is a little (Delphi Pascal) computer program in
one of the answers.
At the end of that piece of code we find:


begin
  anatoly(2,13); { at double precision }
end.

Replace this by:


begin
  anatoly(4,13); { at double precision }
end.

To get:


 4.00000000000000E+0000  2.40161552602630E+0000 +/- 1.80950091432944E-0018

Which is - no big surprise - in concordance with the numerical value as given by Gottfried Helms
in his answer .
