Convergence rate for $a_{n+1}=\sqrt{2 \sqrt {a_n}}$
I don't know the next step after :$$\frac {a_{n+1}^4}{ a_n}={4}$$
Edit, maybe there is no simple answer for this, is there a rate of convergence known for $a_{n+1}=\sqrt{2 {a_n}}$ aka $$\sqrt {2 \sqrt {2 \sqrt {2 \cdots}}}$$
If $f(x) = \sqrt{2}x^{1/4}$, then our sequence is defined by $a_{n+1} = f(a_n)$.
First of all, we need to find the fixed point(s) of the function. In order to do so, it suffices to solve $f(x) = x$. We find $$ f(x) = x \implies\\ x - \sqrt{2}x^{1/4} = 0 \\ x^{1/4}(x^{3/4} - \sqrt{2})= 0 $$ so, $a_n$ will converge to $0$ or $2^{2/3}$, depending on the initial input.
Now, note that $$ f'(x) = \frac{\sqrt{2}}{4}x^{-3/4} $$ In particular, we find that $0 < |f'(2^{2/3})| = 1/4 < 1$. This necessarily means that, where $a_n$ converges to $2^{2/3}$, it converges linearly with $$ \lim_{n \to \infty} \frac{a_{n+1} - 1}{a_n - 1} = \frac {1}{4} $$ On the other hand, $\lim_{x \to 0} |f'(x)| > 1$, which is enough to tell us that $0$ is an unstable limit of this iteration. That is, we will only have a sequence converging to $0$ if $a_1 = 0$.
One question worth asking is "for which positive numbers will $a_n$ converge to $1$?". I'm fairly certain the answer is "all of them".
At the very least, we can note that $0 < f'(x) < 1$ whenever $x > 1/4$, so if we begin with any $a_0 > 1/4$ we have guaranteed convergence. Moreover, because $f'(x) > 0$, this convergence will be monotonic from either direction.
Hint $$a_n =\sqrt {2 \sqrt {2 \sqrt {2 \cdots \sqrt{a_n}}}}=2^{\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^{n-1}}} \sqrt[2^n]{a_1}=2^{1-\frac{1}{2^{n-1}}}\sqrt[2^n]{a_1} $$
Thus $$a_n=2 \hspace{.2cm} \sqrt[2^{n-1}]{\frac{\sqrt{a_1}}{2}}$$
