1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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. $\qquad$ $\endgroup$ – Michael Hardy Jan 24 '16 at 23:29
1
$\begingroup$

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?

$\endgroup$
0
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.