Math Stack Exchange!
I am trying to figure out how to find value of the infinitely nested radical
$$x = \sqrt{2^0 + \sqrt{2^2 + \sqrt{2^4 + \sqrt{2^8 + \ldots}}}}$$
I have already established that this expression converges based on Herschfeld's Convergence Theorem, but I am now interested in proving the specific value to which it converges.
Here is the proof that it converges: Let $a_n = 2^{2^n}$. Then we have $$ a_n^{2^{-n}} = (2^{2^n})^{2^{-n}} = 2^{2^n \cdot 2^{-n}} = 2^1 = 2 $$
In other words, for all $n$, $ a_n^{2^{-n}} = 2 $, which means the supremum of $ a_n^{2^{-n}} $ is 2 by Herschfeld's Convergence Theorem.
Thank you for any help!!
$x^2=1+\sqrt{2^2+\sqrt{2^4+\sqrt{2^8+...}}}=1+2\sqrt{1+\sqrt{1+\sqrt{1+...}}}=1+2u$
Now notice that $u^2=1+u$ so by solving this this quadratic you get $u=\frac{1+\sqrt{5}}{2}$ and finally $x^2=1+2\frac{1+\sqrt{5}}{2}=2+\sqrt{5}$
so $x=\sqrt{2+\sqrt{5}}\approx2.06$