Does the sequence $x_{n+1}=2\sqrt{x_n}$ converge?

Question

Does the sequence $(x_n)_{n=1}^\infty$ with $x_{n+1}=2\sqrt{x_n}$ converge?

I'm almost positive this converges but I am not entirely sure how to go about this. The square root is really throwing me off as I haven't dealt with it at all up until now.

Answer 1

First, find the limit.

Suppose the sequence has a limit L. Then $L=2\sqrt{L}$, so $L=4$. Note: this does not show that the limit exists. It shows that if the limit does exist then it must be 4.

Then, see how the limit is approached.

If $y=2\sqrt{x}$, then $y-4=2\sqrt{x}-4 =2(\sqrt{x}-2) $ so $\frac{y-4}{x-4} =\frac{2(\sqrt{x}-2)}{x-4} =\frac{2(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)} =\frac{2}{\sqrt{x}+2} $.

Therefore $|\frac{y-4}{x-4}| < 1$ so the iterations converge to the limit.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} x_{n + 1} & = 2\root{x_{n}} = 2\, x_{n}^{1/2} = 2^{1 + 1/2}\,x_{n - 1}^{1/2^{2}} = 2^{1 + 1/2 + 1/2^{2}}\, x_{n - 2}^{1/2^{3}} = \cdots = 2^{\large\sum_{k = 0}^{n}2^{-k}}\,x_{0}^{1/2^{\large n + 1}} \\[5mm] & =2^{2 - 2^{-n}}\,x_{0}^{1/2^{\large n + 1}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\, 2^{2} = \bbx{4} \end{align}