$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{x_{0}\equiv{M \over 2},\quad x_{n+1}\equiv\half\,\pars{x_{n} + {M \over x_{n}}}.\qquad M \geq 0.}$
Let's $\ds{y_{n} = {x_{n} \over \root{M}}\quad\imp\quad y_{n + 1} = \half\pars{y_{n} + {1 \over y_{n}}}\,,\quad y_{0} = {\root{M} \over 2}}$ \begin{align} {y_{n + 1} + 1 \over y_{n + 1} - 1}& ={\half\pars{y_{n} + 1/y_{n}} + 1 \over \half\pars{y_{n} + 1/y_{n}} - 1} ={y_{n}^{2} + 2y_{n} + 1 \over y_{n}^{2} - 2y_{n} + 1} =\pars{y_{n} + 1 \over y_{n} - 1}^{2} \end{align}
\begin{align} {y_{n + 1} + 1 \over y_{n + 1} - 1}& =\pars{y_{n - 1} + 1 \over y_{n - 1} - 1}^{4} =\pars{y_{n - 2} + 1 \over y_{n - 2} - 1}^{8}=\cdots =\pars{y_{0} + 1 \over y_{0} - 1}^{2^{n + 1}} \end{align}
$ y_{n + 1} = {\pars{y_{0} + 1}^{2^{n + 1}} + \pars{y_{0} - 1}^{2^{n + 1}} \over \pars{y_{0} + 1}^{2^{n + 1}} - \pars{y_{0} - 1}^{2^{n + 1}}} $ Now, you can take from here.