Recursive radiacls

$3=\sqrt{4+5} = \sqrt{4+\sqrt{16+9}}= \sqrt{4+\sqrt{16+\sqrt{81}}} =\sqrt{4+\sqrt{16+\sqrt{64+17}}}$ So , $3= \sqrt{4+\sqrt{16+\sqrt{64+\sqrt{256+33\cdots}}}}$

This is one of the Ramanujan-style infinite radicals. Consider the sequence of functions $g_m\,:\, [0,\infty) \to [0,\infty)$ defined inductively as follows: $g_0(c) \; = \; c \hspace{2cm} g_m(c) \; = \; \sqrt{c^2 + g_{m-1}(2c)} \;\hspace{2.5cm} c \ge 0\,,\,m \in \mathbb{N}$ It is clear that $g_m(c) \ge c$ for all $c \ge 0$ and $m \ge 0$ . Certainly $g_o(c) \le c$ for all $c \ge 0$ . If $g_{m-1}(c) \le c+1$ for all $c \ge 0$ , it follows that $g_m(c)^2 \le c^2 + (2c+1) = (c+1)^2$ , so that $g_m(c) \le c+1$ for all $c \ge 0$ . Thus we deduce by induction that $c \;\le\; g_m(c) \;\le \; c+1 \hspace{2cm} c \ge 0\,,\, m \ge 0$ Since $0 \le c+1 - g_m(c) \; = \; \frac{(c+1)^2 - g_m(c)^2}{c+1+g_m(c)} \; \le \; \frac{(c+1)^2 - c^2 - g_{m-1}(2c)}{c + 1 + g_m(c)} \; \le \; \frac{2c+1 - g_{m-1}(2c)}{2c+1}$ for all $c \ge 0$ and $m \ge 1$ , we deduce by induction that $0 \le c+1 - g_m(c) \; \le \; \frac{2^kc + 1 - g_{m-k}(2^kc)}{(2c+1)(4c+1) ... (2^kc+1)} \hspace{2cm} c \ge 0\,,\, 1 \le k \le m$ so that $0 \le c+1 - g_m(c) \; \le \; \frac{2^mc + 1 - g_0(2^mc)}{(2c+1)(4c+1) ... (2^mc+1)} \; \le \; \frac{1}{2^mc + 1} \hspace{2cm} c \ge 0\,,\, m \ge 0$ and hence $\lim_{m \to \infty} g_m(c) \; = \; c + 1 \hspace{2cm} c > 0$ In this case we have $c=2$ , and hence the infinite radical equals $\boxed{3}$ .