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

$$\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+...}}}}=?$$ I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to 2. But how can this problem be solved with proper procedure?

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
$$F_n<2 Hence $$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}+...}}}}$$