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I want to prove that the sequence given by:

$$x_0=1$$ $$x_{n+1}=\frac{1}{2}\left( x_n+ \frac{2}{x_n}\right)$$

is a Cauchy sequence. In order to do this, my approach was to prove that it is a convergent sequence and the conclude it must be a Cauchy sequence. I know that I can say that the sequence is monotone and bounded and hence it converges (As they do here Convergence of $x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right).$), but I am interest in doing the proof using the definition of convergence . Namely: $x_n$ converges to $x$ if $\forall \epsilon$, $\exists N \in \mathbb{N}$ such that if $n>N$ then $|x_n-x|<\epsilon$.

So far I haven't succeded finding the right $N$. Any suggestions?

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Weierstraß Ramirez
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  • If you want to show this converges by using the definition limit of a sequence then why does your subject line say you want to prove it's a Cauchy sequence? You can show it converges by showing it's a Cauchy sequence and then citing a theorem that says that Cauchy sequences converge, but that's not the same as directly using the definition of convergence. $\qquad$ – Michael Hardy Jan 24 '16 at 23:29

2 Answers2

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Hint: first you find $x$ by solving : $x = \dfrac{x+\dfrac{2}{x}}{2}$. Thus ...$x = \sqrt{2}$. Next you write: $|x_{n+1} - \sqrt{2}| = \dfrac{(x_{n}-\sqrt{2})^2}{2x_n} \leq \dfrac{(x_n-\sqrt{2})^2}{2}$ since $x_n \geq 1, \forall n \geq 0$..Can you continue?

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$\forall \epsilon$, Let N = max{ 1, $\lfloor \frac{x_0-\sqrt{2}}{\epsilon}\rfloor$+1} , when n > N
$0 \le |x_{n+1} - \sqrt{2}| = \frac{(x_n - \sqrt{2})^2}{2x_n} \le \frac{(x_0 - \sqrt{2})}{2^n} < \frac{(x_0 - \sqrt{2})}{n} < \epsilon$

DeepSea
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