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I struggle to prove this using the definition only. I can prove it by using some limit laws, but I wanted to use only the definition out of curiosity. I would appreciate some hints.

Q: Prove that the sequnece below converges to $\sqrt{2}$: $$x_{n+1}=\frac{1}{2}x_n+\frac{1}{x_n},$$ where $x_1 = 2$.

chuck
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    Use induction to prove $x_{n+1}\leq x_n$ and $x_n>\sqrt{2}$ to conclude that a the limit exists and then solve $x=\frac{x}{2}+\frac{1}{x}$ to find the limit $x.$ – Frieder Jäckel Sep 12 '17 at 23:51
  • No need to show that $x_n > \sqrt{2}$. It is clear that $x_n \ge 0$. – velut luna Sep 12 '17 at 23:54
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    @velutluna You need the monotonicity, and you need $x_n>\sqrt{2}$ to get the monotonicity. In particular, $\frac{1}{2} + \frac{1}{1}>1$. (On the other hand it is also bigger than $\sqrt{2}$ now, so such initial conditions will also converge to $\sqrt{2}$. Proving that just requires an additional step that isn't needed in this problem.) – Ian Sep 12 '17 at 23:58
  • @Ian Provided that monotonicity is proved, then only need to show that it's bounded from below. Isn't it? – velut luna Sep 13 '17 at 00:00
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    @velutluna Strictly speaking yes, but $x_{n+1} \leq x_n$ is is not true if $x_n \in (0,\sqrt{2})$. So you will be unable to prove monotonicity without this stronger lower bound. – Ian Sep 13 '17 at 00:01
  • @Ian Yes you are right. – velut luna Sep 13 '17 at 00:02
  • @Ian I'm not sure if you noticed, but $x_1=2\neq 1$, so it should be $\frac{1}{2}\cdot 2+\frac{1}{2}$, which still equals $\frac{1}{2}+\frac{1}{1}$. – user263326 Sep 13 '17 at 00:08
  • @user263326 I was considering what happens when you plug in $x_1=1$ (which is less than $\sqrt{2}$) instead. In this case you get a value bigger than what you put in, which violates the monotonicity that you want here. – Ian Sep 13 '17 at 00:44

2 Answers2

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We have some observations about $x_n$:

$1)$ $x_{n+1} - x_n = \dfrac{1}{x_n} - \dfrac{x_n}{2}= \dfrac{2-x_n^2}{2x_n} \le 0$.

$2)$ $x_{n+1} \ge \sqrt{2}\implies x_{n+1}^2 \ge 2$ for all $n \ge 1$, by AM-GM inequality.

These two observations imply that the limit $L$ exists and $L = \dfrac{L}{2} + \dfrac{1}{L}$. This gives: $L = \sqrt{2}$.

DeepSea
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    You should switch $1)$, $2)$, because $1)$ uses $2)$. Also, you need to prove $x_n>0$ before you can use AM-GM inequality. – user263326 Sep 13 '17 at 00:11
  • $x_n>0$ is trivial by $x_1>0$ and induction. – user263326 Sep 13 '17 at 00:57
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    You don't even separately argue $x_n>0$, the inductive hypothesis for part 2 is $x_n \geq \sqrt{2}$ already. (Incidentally, part 2 should really be part 1, because you need it for part 1.) – Ian Sep 13 '17 at 01:02
  • DeepSea: Why do you regularly fail to address the serious objections made about your "answers"? – Did Sep 13 '17 at 09:16
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Under the substitution $x_n:=\sqrt{2}+y_n$ $\ (n\geq1)$ the given recursion rewrites to $$y_1=2-\sqrt{2}\doteq0.586,\qquad y_{n+1}={y_n^2\over 2(\sqrt{2}+y_n)}\qquad(n\geq1)\ .$$ It is then obvious that $\lim_{n\to\infty}y_n=0$.

Christian Blatter
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