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Let be $(x_n)$ a sequence defined by $x_1=1$ and $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$.

$(x_n)$ is increasing or decreasing? I do not know if $(x_n)$ is bounded. I would like to know if $(x_n)$ is convergent. Any suggestions?

Bobson Dugnutt
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Nelly
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1 Answers1

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Oups! I thought that the post was saying that the sequence in increasing... Sorry for this. To make things clear I will provide full solution.

We will show that $(x_n)$ is ${\bf decreasing}$ and bounded from below by $\sqrt{2}$ from the second term and after and that $\lim_n x_n=\sqrt{2}.$

$x_1=1,$ $x_2=\frac{3}{2}>\sqrt{2}.$ Let $x_n\geq \sqrt{2}$ for some $n\geq 2$ (induction hypothesis).

Then $x_{n+1}\geq \sqrt{2}\;\Leftrightarrow\;x_{n+1}^2\geq 2\;\Leftrightarrow\;(x_n+\frac{2}{x_n})^2\geq 8\;\Leftrightarrow\;(x_n-\frac{2}{x_n})^2\geq 0,$ which holds.

We have showed that $x_n\geq \sqrt{2}\;\forall\;n\geq 2.$

Let $n\geq 2.$ $x_{n+1}\leq x_n\;\Leftrightarrow\;\frac{1}{2}(x_n+\frac{2}{x_n})\leq x_n\;\Leftrightarrow\;\frac{2}{x_n}\leq x_n\;\Leftrightarrow\;2\leq x_n^2,$ which is true for any $n\geq 2.$

So, $(x_n)$ is convergent to $\inf_n x_n\geq \sqrt{2}.$

If $\lim_n x_{n+1}=\lim_n x_n=x,$ the inductive formula gives $x^2=2,$ hence $x=\sqrt{2},$ since $x>0.$

This is actually a formula due to Newton in order to approximate (from above) the number $\sqrt{2}.$ More generally, for any $\alpha>0,$ if you set $x_1$ to be any positive number, say $x_1=1,$ and then $x_{n+1}=\frac{1}{2}(x_n+\frac{a}{x_n}),$ for any $n\in\mathbb{N},$ you can show that for $n\geq 2$ the sequence in decreasing and that it is bounded from below by $\sqrt{a},$ hence converges (by the same reasoning, to $\sqrt{a}).$

Andreas K.
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    Actually $x_n$ isn't increasing... – Wojowu Mar 10 '16 at 18:09
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    No, you can't show that the sequence is increasing or decreasing by assuming it has a limit. – David C. Ullrich Mar 10 '16 at 18:39
  • "Now pick any number $\alpha \geq \sqrt{2}$ and show (easily with induction) that $x_n\leq \alpha$ for any $n\in\mathbb{N}.$" Sorry but there exists some $\alpha>\sqrt2$ such that this is not true (and this even less true for $\alpha=\sqrt2$...). – Did Mar 10 '16 at 20:27
  • @Wojowu Yes, I edited. I thought that OP had posted that he had already proved that the sequence was decreasing. – Andreas K. Mar 12 '16 at 04:20
  • @David Yes, you can do it (at some cases!). Note that if you have an inductive formula of a sequence and you show that the sequence is monotone, then if you ${\bf assume}$ that it converges, say to $x$ and you can find $x$ from the inductive formula, then you can show that the sequence is bounded from below (if the sequence is decreasing) or from above (if the sequence is increasing) by $x.$ Remember that a convergent increasing sequence (decreasing resp.) converges to the $\sup$ of the terms ($\inf$ of the terms resp.). – Andreas K. Mar 12 '16 at 04:24
  • @Did. See the updated post. What I initially wrote is true for any $\alpha\leq \sqrt{2}.$ – Andreas K. Mar 12 '16 at 04:26
  • No, "what you initially wrote" is not true either "for any $\alpha\leqslant\sqrt2$"... but this is no big deal since you now entirely revised the answer. – Did Mar 12 '16 at 07:16
  • Your comment "if ... you show that the sequence is monotone, then if you assume that it converges, say to $x$ and you can find $x$ from the inductive formula, then you can show that the sequence is bounded from below" has absolutely no relevance to what I said, that you can't show a sequence is monotone by assuming it has a limit. – David C. Ullrich Mar 12 '16 at 13:37