Determine $x$ such that $\lim\limits_{n\to\infty} \sqrt{1+\sqrt{x+\sqrt{x^2…+\sqrt{x^n}}}} = 2$

Solution 1:

Let me describe a sketch of proof that $x=4$.

A. Observe that if $f(x)=\lim_{n\to\infty}\sqrt{1+\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n}}}}$, then $f$ is strictly increasing.

B. We shall show that $f(4)=2$, and hence $x=4$ is the unique answer.

$B_1.$ Fix $m\in\mathbb N$ and show that, for $n=m,m-1,m-2,\cdots$ (induction backwards) $$ 2^n<\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}<2^n+1, $$ while $$ \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}=2^n+1. $$

$B_2.$ Next estimate the difference $$ (2^n+1)- \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\ =\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}- \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}} \\ =\frac{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}- \sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}{\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}}+ \sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}} \\ <\frac{{\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}+1}}- \sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}}{2\cdot 2^n} \\ <\cdots<\frac{(\sqrt{4^m}+1)-\sqrt{4^m}}{2^{m-n}\cdots 2^{n+(n+1)+\cdots+(m-1)}}=2^{-\frac{(m-n)(n+m+1)}{2}} $$ Thus $$ \lim_{m\to\infty}\sqrt{4^n+\sqrt{4^{n+1}+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^n+1. $$ For $n=0$ we have $$ \lim_{m\to\infty}\sqrt{1+\sqrt{4+\cdots\sqrt{4^{m-1}+\sqrt{4^m}}}}=2^0+1=2. $$

Solution 2:

Hint: Pretend that there's an extra $+1$ at the next-to-last level. Then $$4^{n-1}+\sqrt{4^n}+1~=~2^{2(n-1)}+2^n+1~=~2^{2(n-1)}+2\cdot2^{n-1}+1~=~(2^{n-1}+1)^2.$$ Can you see what happens ? :-$)~$ Now, as $n\to\infty,$ the numerical influence gained by adding that extra $+1$ at the top level tends towards $0.$