Bounded sequence $a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$ Let
$$a_n=\sqrt{4+2 \sqrt{4+\cdots+2 \sqrt{4+2 \sqrt{4+2 \sqrt{4+4}}}}}$$
the sign $\sqrt{}$ occurs $n$ times.
a) Prove, that $a_n< \sqrt{5}+1$ for all $n$.
b) Find $\lim_{n\rightarrow \infty } a_n$
Author O.Kukush

You can prove (a) by mathematical induction. $a_1$ is less than $1+\sqrt5$ by direct comparison as $a_1 = \sqrt8 < 3$ while $1+\sqrt5>3$. Let $a_n < 1+\sqrt5$. Then $4+2a_n < 6+2\sqrt5$ and taking square root on both sides (that both are positive is trivial) we obtain $a_{n+1} < 1+\sqrt5$.
It also follows that $(a_n)$ is strictly growing. You can check that by comparing $a_{n+1}^2 = 4+2a_n$ with $a_n^2$ knowing that (a) is true.
From the monotonicity and boundedness it follows that $(a_n)$ has a limit and this number, let's call it $a$, needs to be a fixed point of $\sqrt{4+2a} = a$. The only solution larger than $0$ is $1+\sqrt5$.
