Proving recursive sequence converges to $\sqrt{2}$

Question

Consider the sequence $x_{n+1} = \frac{1}{2} (x_n + \frac{2}{x_n})$, $x_1 = 2$. Prove that it converges to $\sqrt{2}$.

I want to show that all of $x_n$ is bounded below by $\sqrt{2}$ using induction. However, I can't see how knowing $x_n \geq \sqrt{2}$ helps show that $\frac{1}{2} (x_n + \frac{2}{x_n}) \geq \sqrt{2}$. Namely, it's not clear whether a higher value of $x_n$ produces a lower (or higher) value of $x_{n+1}$.

Edit: I am looking for a solution that does not use AM-GM (which is proven later in the text that I am reading).

I believe your sequence is monotone decreasing...Sonnhard Grauber's suggestion to use AM-GM seems like a great approach... — User001, Apr 09 '16 at 05:31
I'm pretty sure that this is a duplicate, but I can't find it right now. — marty cohen, Apr 09 '16 at 05:45
See also Proof of Convergence: Babylonian Method $x_{n+1}=\frac{1}{2}(x_n + \frac{a}{x_n})$ (and maybe other posts linked there.) — Martin Sleziak, Apr 09 '16 at 06:45
@SubhadeepDey You see the list in the linked questions in the sidebar, right? If there is too many of them to fit there, you can click on the link "see more linked questions..." The url has format as above. (You can test this basically on any question from the top of the list in the frequent tab.) — Martin Sleziak, Apr 09 '16 at 06:54
Thanks for the suggestions. I found this link most helpful: https://proofwiki.org/wiki/Iterative_Process_for_Estimating_Square_Roots — rorty, Apr 09 '16 at 07:19

score 1 · Answer 1 · answered Apr 09 '16 at 06:06

First I would like to point out that it is possible to find a duplicate for this question because it has been asked quite a few times here. So I am trying to give a fresher approach based on "completing square". Observe that:

$1)$ $x_n > 0, \forall n \geq 1$.

$2)$ $x_{n+1} - 1 = \dfrac{x_n+\dfrac{2}{x_n}}{2}-1=\dfrac{(x_n-1)^2+1}{2x_n}> 0\Rightarrow x_n > 1, \forall n \geq 1$.

$3)$$0 \leq |x_{n+1} - \sqrt{2}| = \dfrac{|x_n-\sqrt{2}|^2}{2x_n}< \dfrac{|x_n-\sqrt{2}|^2}{2} \leq|x_n - \sqrt{2}|^2\Rightarrow |x_{n+1} -\sqrt{2}|< |x_1-\sqrt{2}|^{2^n}=|2-\sqrt{2}|^{2^n}$, and apply squeeze lemma to get the limit of $\sqrt{2}$ as claimed.

score 0 · Answer 2 · answered Apr 09 '16 at 05:31

0

prove by induction that $$x_n>0$$ for all natural $$n$$ and then we get by AM-GM: $$x_{n+1}=\frac{1}{2}(x_n+\frac{1}{x_n})\geq\sqrt{2}$$

answered Apr 09 '16 at 05:31

Dr. Sonnhard Graubner

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Proving recursive sequence converges to $\sqrt{2}$

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