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Given: $\sqrt{\alpha}<x_1, \alpha >0, x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$. Show $\lim_{n\to \infty}x_n=\sqrt{\alpha}.$

I have already shown the this is a monotonic decreasing function. Intuitively, I see in the end I get $\frac{1}{2}\left(\sqrt{\alpha}+\sqrt{\alpha}\right)$ but I'm not sure if I should keep going using $\lim$ notation or if I should switch to the definition of converging sequence.

cheeseboardqueen
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  • This has probably been asked here before, but can you show it's monotonic decreasing and bounded? Then it converges to $\lambda$ where $\lambda=\frac12(\lambda+\frac\alpha\lambda)$; i.e., $\lambda^2=\alpha$ – J. W. Tanner Dec 14 '20 at 03:26
  • @J.W.Tanner Oh! I can say it's bounded above by $x_1$. Am I allowed to say ${x_n}$ is bounded below (?) because I found each $x_n>\sqrt{\alpha}$ so that should imply $\sqrt{\alpha}$ is the lower bound. Then as the sequence is monotonic and decreasing and bounded, then I basically have it for free. – cheeseboardqueen Dec 14 '20 at 03:31
  • if it's monotonic decreasing and bounded below (say by $\sqrt\alpha$), then it's convergent – J. W. Tanner Dec 14 '20 at 03:47
  • Cf. this question – J. W. Tanner Dec 14 '20 at 03:56

3 Answers3

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Rewrite $$ x_{n+1}=\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)$$ $$ x_{n+1}= x_{n}-\Big[x_n-\frac{1}{2}\left(x_n+\frac{\alpha}{x_n}\right)\Big]$$ that is to say $$ x_{n+1}= x_{n}-\frac {x_n^2-\alpha}{2x_n}$$ and recognize the Newton iterative scheme for finding the zero of function $f(x)=x^2-\alpha$.

Claude Leibovici
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Since this hasn't been closed as duplicate, here's my answer again.

A more direct proof.

$s_{n+1} =\dfrac12(s_n+\dfrac{a}{s_n}) $.

$\begin{array}\\ s_{n+1}^2 &=\dfrac14(a_n^2+2a+\dfrac{a^2}{s_n^2})\\ &=a+\dfrac14(s_n^2-2a+\dfrac{a^2}{s_n^2})\\ &=a+\dfrac14(s_n-\dfrac{a}{s_n})^2\\ &\gt a\\ \end{array} $

so $s_n > \sqrt{a} $.

$\begin{array}\\ s_{n+1}-\sqrt{a} &=\dfrac12(s_n-2\sqrt{a}+\dfrac{a}{s_n})\\ &=\dfrac12(\sqrt{s_n}-\dfrac{\sqrt{a}}{\sqrt{s_n}})^2\\ &=\dfrac1{2s_n}(s_n-\sqrt{a})^2\\ \dfrac{s_{n+1}-\sqrt{a}}{s_n-\sqrt{a}} &=\dfrac1{2s_n}(s_n-\sqrt{a})\\ &\lt \dfrac12\\ \end{array} $

so $s_n$ converges to $\sqrt{a}$.

marty cohen
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The problem is well-known here of course and in most intro to analysis books for sure. You probably come across few proofs or stumbled upon a page in some book that talks about this or similar problem. I just wanna write up what I see as perhaps what you want to see:

-For all $n\ge 1, x_n \ge \sqrt{\alpha}$. This fact can be established by the AM-GM inequality.

-The sequence is monotonically decreasing: $x_{n+1} - x_n = \dfrac{1}{2}\left(x_n+\dfrac{\alpha}{x_n}\right)-x_n=\dfrac{1}{2x_n}\left(\alpha - x_n^2\right)\le 0$

The above information shows that the sequence is bounded below and decreasing, hence converges to a limit $L$. You can show the $L = \sqrt{\alpha}$ from the equation defining $x_n$.