Infintely nested radical

Question

Recently, I saw this intriguing radical, which is infinitely nested. I tried to calculate its value by considering the term $(2^n)^2$ inside $(n+1)$ square roots but could not de-nest it. But by using the concept of an approaching (limit) method, its value seems $2$. Could you please provide a method to derive it? $\displaystyle\sqrt{1^2+\sqrt{2^2+\sqrt{4^2+\sqrt{8^2+\sqrt{16^2+\sqrt{32^2+\cdots}}}}}}=2$

Answer 1

Let $$a_n = \sqrt{1^2+\sqrt{2^2+\sqrt{4^2+…+\sqrt{2^{2n}}}}}.$$ We see that the sequence $(a_n)$ is increasing. Also we have:

$$a_n = \sqrt{1^2+\sqrt{2^2+…+\sqrt{2^{2n-2} +\sqrt{2^{2n}}}}} =$$ $$= \sqrt{1^2+\sqrt{2^2+…+\sqrt{2^{2n-2} + 2^n}}} \le $$

$$\le \sqrt{1^2+\sqrt{2^2+…+\sqrt{2^{2n-2} + 2\cdot 2^{n-1}+1}}} =$$ $$= \sqrt{1^2+\sqrt{2^2+…+\sqrt{(2^{n-1}+1)^2}}} = $$

$$ =\sqrt{1^2+\sqrt{2^2+…+\sqrt{2^{2n-4} + 2^{n-1}+1}}}=$$ $$=\sqrt{1^2+\sqrt{2^2+…+\sqrt{2^{2n-6} + 2^{n-2}+1}}}=$$

$$=… =\sqrt{1^2+ 2\cdot 1+1}=2$$

So, $a_n$ is increasing and bounded, then it converges to some $a$. We want to prove that $a=2.$ $a_n \sim 2$ is equivalent to $$2^{2n}\sim (…((((2^2-1^2)^2-2^2)^2-4^2)^2-8^2)^2-…-2^{2n-2})^2.$$

It’s easy to show by induction that the RHS $=(2^{n}+1)^2$. Indeed, $$((2^{n}+1)^2-2^{2n})^2=(2^{2n}+2\cdot2^n+1-2^{2n})^2= (2^{n+1}+1)^2.$$

And since $2^{2n}\sim (2^{n}+1)^2$, we’re done.