<s>[INST] The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$ 
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$

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?
 [/INST] 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.
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