<s>[INST] Convergence of an infinite sequence 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!
 [/INST] 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.

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.
</s>