<s>[INST] 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)
 [/INST] 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.
</s>