Let $a_{1}=\sqrt{2}$, and consider the sequence of positive real numbers defined by

$$a_{n+1}=\sqrt{2+\sqrt{a}_{n}}, \quad n=1,2,3, \ldots$$

Show that $a_{n} \leqslant 2$ for all $n$. Prove that the sequence $a_{1}, a_{2}, \ldots$ converges to a limit.

Suppose instead that $a_{1}=4$. Prove that again the sequence $a_{1}, a_{2}, \ldots$ converges to a limit.

Prove that the limits obtained in the two cases are equal.