Convergence of an infinite sequence

Question

I want to know if
$$ B=\sqrt{2+\sqrt{4+\sqrt{8+\sqrt{16+...}}}} $$ converges to a finite value or not. $B$ can be written as $$ B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+...}}}}. $$ However, I have no idea the next step. Any suggestion, idea, or comment is welcome, thanks!

Answer 1

We have

$$B=\sqrt{2^1+\sqrt{2^2+\sqrt{2^3+\sqrt{2^4+\cdots}}}}<\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}$$

We then note that

$$A=\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}+\sqrt{2^{2^3}+\cdots}}}}\\A=\sqrt{2+A\sqrt2}\implies A=\frac{\sqrt2+\sqrt{10}}2$$

Thus, $B$ is bounded above, so it converges.

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One may see $A$ is bounded above since

$$A_1=\sqrt{2^{2^0}}=\sqrt2=\frac{\sqrt2+\sqrt2}2<\frac{\sqrt2+\sqrt{10}}2$$

$$A_2=\sqrt{2^{2^0}+\sqrt{2^{2^1}}}=\sqrt{2+\sqrt2A_1}<\sqrt{2+\sqrt2\frac{\sqrt2+\sqrt{10}}2}=\frac{\sqrt2+\sqrt{10}}2$$

$$A_3=\sqrt{2^{2^0}+\sqrt{2^{2^1}+\sqrt{2^{2^2}}}}=\sqrt{2+\sqrt2A_2}<\sqrt{2+\sqrt2\frac{\sqrt2+\sqrt{10}}2}=\frac{\sqrt2+\sqrt{10}}2$$

And on with the induction.