# Prove that $x_n$ converges, and find $\lim_{n\rightarrow \infty} x_n$

I need some critique and suggestions on how to go over this question:

Prove that $x_n$ converges, and find $\lim_{n\rightarrow \infty} x_n$:

$x_1 > 0$, $x_{n+1} = \frac{1}{2}(x_n + \frac{5}{x_n}) \ \forall n \geq 1$.

Here's how I solved it:

Let $x_1 =1 \Rightarrow x_2 = 3 \Rightarrow x_3 = \frac{28}{12} < x_2$ and $x_4 < x_3$.

So the sequence is actually decreasing and therefore it is bounded below by $x_1$.

Let $\lim_{n\rightarrow \infty} x_{n+1} = \lim_{n\rightarrow \infty} x_{n} = A$.

$A =\frac{1}{2} (A+\frac{5}{A}) \Rightarrow A = \sqrt5$.

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Showing that $x_4\lt x_3$ doesn't show the sequence decreases for all $x_n$. –  apnorton Oct 8 '13 at 0:04
That's where I stucked at. The sequence is very contractile from $x_6$ on. So probably I should mention the contractive principle in the proof ? –  Logarithm Oct 8 '13 at 0:10
I don't think that's necessary, Logarithm. I believe anorton is attempting to imply that you need to utilize induction on the proof. Don't make it too complicated. If it's monotonic and bounded, then it is convergent by definition. No need to use anything more complicated than that. –  user99464 Oct 8 '13 at 0:16

You have the right idea. You have a recursive formula and to show it has a limit, first show it is bounded--this can be done by induction, then show it is monotone, and then yes it correct to assert that $lim$ $a_{n+1}$ = $lim$ $a_n$.

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$$x_{n+1} - \sqrt{5} = \frac{1}{2}\cdot\frac{(x_n-\sqrt{5})^2}{x_n}$$

so if $x_0 \ge \sqrt{5}$ then with the equation above $x_n \ge \sqrt{5}$ then $$x_{n+1}-\sqrt{5} \le \frac{1}{2\cdot\sqrt{5}}(x_n-\sqrt{5})^2$$

then $$x_{n}-\sqrt{5} \le \left(\frac{1}{2\cdot\sqrt{5}}\right)^{2^{n}-1} (x_0-\sqrt{5})^2$$

thus $x_n \to \sqrt{5}$

actually this sequence is used to approximate the root of 5, and it's simply the newton-raphson algorithm fo the function $f(x) = \frac{1}{2}(x+\frac{5}{x})$

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The computer is right. But you have to show that $\lim_{n\to \infty}x_n$ exists at first.

First, $x_{n+1}\geq \sqrt{5}$. Then $x_{n+1}-x_n=\frac{1}{2}\frac{5-x_n^2}{x_n}\leq 0$, and $\{x_n\}$ is decreasing. Therefore, the limit $\lim_n x_n$ exits.

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