The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$

Note that: $2^2=1+3$, $3^2=4+5$, $5^2=16+9$, $9^2=64+17$, ...

Therefore $$2=\sqrt{4^0+3}$$ $$2=\sqrt{4^0+\sqrt{4^1+ 5}}$$ $$...$$ $$2=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+17}}}}$$ $$...$$ $$...$$

Let $F_n=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+...}}}}$ where the sequence terminates after $n$ square roots. For positive numbers $a$ and $b$, we have $\sqrt{a+b}<\sqrt{a}+\frac{b}{2\sqrt{a}}$ and therefore

$$F_n<2<F_n+\frac{2^n+1}{2^n(1+2+...+2^{n-1})}$$ Hence $F_n$ converges to 2.

Take $f(x,n)=x+2^n$. We can see that; $$\begin{aligned} f(x,n) &= \sqrt{2^{2n}+x\left(x+2^{n+1}\right)} \\ &= \sqrt{2^{2n}+xf(x,n+1)} \\ &= \sqrt{2^{2n}+x\sqrt{2^{2\left(n+1\right)}+x\sqrt{2^{2\left(n+2\right)}+x\sqrt{...}}}}\\ &=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\\ \end{aligned}$$

Taking $x=1,n=0$; we get; $$2=\sqrt{4^{0}+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}$$