Be $a>0$ a real number. Define by the induction method the sequence $x_{n+1}= \frac{[x_n + \frac{a}{x_n}]}{2}$.
Show that the sequence $(x_n)$ thus obtained converges to $\sqrt{a}.$
\begin{align*}
\forall x \in \mathbb{R} &\Rightarrow \sqrt{a}<x\\ &\Rightarrow \bigg(\frac{a}{x}\bigg)<\sqrt{a}<x \hspace{0.2cm}(\mbox{multiplying both sides of inequality by}\hspace{0.05cm} \sqrt{a})
\end{align*}
\begin{align*}
i)& \hspace{0.8cm} x_{n+1}=\frac{1}{2}\cdot \bigg(x_n+\frac{a}{x_n}\bigg)\\
ii)& \hspace{0.8cm} y=\frac{1}{2}\cdot \bigg(x_n+\frac{a}{x_n}\bigg)
\end{align*}
\begin{eqnarray}
\sqrt{a}<x \Rightarrow a<x\cdot \sqrt{a} \Rightarrow \frac{a}{x}<\sqrt{a}<x. \hspace{2cm}(1) \\
y=\bigg(\frac{x+\frac{a}{x}}{2}\bigg) \stackrel{({1})}{<}\frac{x+x}{2}=x, \hspace{0.5cm}\mbox{i.e}\hspace{0.5cm}(y<x)\hspace{2cm}(2)\\
y=\bigg(\frac{x+\frac{a}{x}}{2}\bigg)\stackrel{({1})}{>}\frac{\sqrt{a}+\sqrt{a}}{2}=\sqrt{a}\hspace{2cm}(3)
\end{eqnarray}
By $(2)$ and $ (3) $ we have
\begin{equation}
\sqrt{a}<y<x \hspace{8.5cm}(4)
\end{equation}
But, I need to prove that
\begin{equation}
x_1>x_2>...x_n>x_{n+1}>... \hspace{5cm}(5)
\end{equation}
Because, for $ (5) $ I get to show that:
$ x_n $ is a sequence monotone limited and convergent.
But, how do I prove this?
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Fernando Sousa
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Your $ a$ is $<0$ and you write $\sqrt{a}$ ? – hamam_Abdallah May 19 '21 at 19:21
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@hamam_Abdallah i did the correction for $a>0$. – Fernando Sousa May 19 '21 at 19:23
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Does $[x_n + \frac{a}{x_n}]$ mean that the values are truncated down to an integer? In that case $(x_n)$ cannot converge to $\sqrt a$ for arbitrary $a> 0$. – Martin R May 19 '21 at 19:40
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If $[...]$ are normal parentheses then you can find many solutions on this site, e.g. here: https://math.stackexchange.com/q/82682/42969. – Martin R May 19 '21 at 19:41