# Asymptotics to the sequence leading up to an infinitely nested radical

Inspired by one of my previous questions, I have found that if $y_0=0$ and $y_{n+1}=\sqrt{2+y_n}$, then

$$y_n=2\cos(\pi/2^n)$$

From there, it's easy to see from Taylor expansions that

$$y_n=2-\frac{\pi^2}{4^n}+\mathcal O(16^{-n})$$

Does it hold true that if $y_0=0$ and $y_{n+1}=\sqrt{a+y_n}$, then

$$y_n\stackrel?=\frac{1+\sqrt{1+4a}}{2}\left(1-\frac{\pi^2}{2\times4^n}+\mathcal O(16^{-n})\right)$$

with $a>0$.

Or perhaps

$$y_n\stackrel?=\frac{1+\sqrt{1+4a}}{2}\left(1-\frac x{r^n}+\mathcal O(r^{-2n})\right)$$

for some constants $x,r$?