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How can we prove that the limit of the sequence $x_n$ defined by: $ x_{n+1}=\frac{1}{2}\big(\frac{x_n+3}{ x_n}\big)$, with $x_0=1$ exists? I have tried to prove that it is Cauchy , but I failed? Can I get some help, and thanks in advance.

Michael Hardy
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Kamal
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    If the limit exists, it has to satisfy the equation $x = \frac 1 2 \left( \frac {x+3} x\right). \qquad$ – Michael Hardy Nov 29 '19 at 23:47
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    with the method of contraction mapping ? :) – Kermatoni Nov 29 '19 at 23:50
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    That implies $2x^2 = x+3.$ So $$x = \frac{1\pm\sqrt{1+24}} 4 = \frac{1\pm 5} 4 = \left( -1 \text{ or } \frac 3 2\right). $$ If you plug positive number into $\frac12\left( \frac {x+3} x\right)$ the result is positive, so the limit cannot be $-1.$ So the problem is to prove that the limit is $3/2. \qquad$ – Michael Hardy Nov 29 '19 at 23:54
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    The terms $x_{2n}$ are monotonically increasing ($x_{2n}<x_{2(n+1)})$ and the terms $x_{2n-1}$ are monotonically decreasing. But fixed point theorem is a much easier way to solve this. – Math1000 Nov 30 '19 at 00:07

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As @Michael Hardy noted, letting $n\to\infty$ in the recurrence relation suggests that $\lim_{n\to\infty} x_n = \frac32$.

We can prove inductively that $x_n \in \left[\frac54,2\right]$ for all $n \ge 1$. Indeed, we have $x_1 = 2$. Furthermore, if $x_n \in \left[\frac54,2\right]$ then also $$\frac54 = \frac12 + \frac32\cdot \frac12\le \underbrace{\frac12 + \frac32 \frac1{x_n}}_{=x_{n+1}} \le \frac12 + \frac32\cdot \frac45 = \frac{11}{10} \le 2$$ so $x_{n+1} \in \left[\frac54,2\right]$.

Now for all $n \ge 1$ have

$$\left|\frac32-x_{n+1}\right| = \left|\frac32-\frac12-\frac32\frac1{x_n}\right| = \left|1-\frac32\frac1{x_n}\right| = \frac1{x_n}\left|x_n - \frac32\right| \le \frac45\left|x_n - \frac32\right|$$

so iterating this yields $$\left|x_n - \frac32\right| \le \left(\frac45\right)^{n-1}\left|x_1 - \frac32\right| = \frac12\left(\frac45\right)^{n-1} \xrightarrow{n\to\infty} 0$$

We conclude that $\lim_{n\to\infty} x_n = \frac32$.

mechanodroid
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Hint : Show that $x_n$'s (for $n\geq 1$) lie in the interval $[\frac 54,2]$ and the function $f(x)=\frac 12+\frac 3{2x}$ on $[\frac 54,2]$ gives a contraction mapping.

Pythagoras
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Just rewriting$$x_{n+1}=\frac{1}{2}\left(\frac{x_n+3}{ x_n}\right)\implies x_{n+1}=x_n-\frac{2 x_n^2-x_n-3}{2 x_n}$$ and recognize the Newton formula for finding the zero's of function $$f(x)=2x^2-x-3$$

Claude Leibovici
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I think with fixed-point theorem : let's define $$ f(x) := \frac{x+3}{2x},$$

such that $x_{n+1}=f(x_n)$.

The interval $[1;2]$ is stable by $f$, so for all $n$, $1 \leqslant x_n \leqslant 2$.

Moreover $f$ is continuous on $[1;2]$ and $3/2$ is the unique fixed point in $[1;2]$, so $(x_n)$ converges to $3/2$.

Kermatoni
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