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)
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.