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

For the case (b) when $n \to \infty$ we can write the following,
\begin{equation}
a_n^2 = 4 + 2a_n \\ a_n^2 - 2a_n - 4 = 0 \\ a_n = \cfrac{2+\sqrt{20}}{2} = 1 + \sqrt{5}
\end{equation}
So, it may be stated that for all other $n$, $a_n < 1 + \sqrt{5} $
