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Sequence $\{x_{n}\}$ is defined by $$x_{1}=1,x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$$, as for proving whether the sequence has a limit, one of my friends told me his proof as following:

First, assuming the sequence has a limit $x$, then take limit on both sides of $$x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$$, that is $$\lim _{x\rightarrow \infty }x_{n+1} =\lim _{x\rightarrow\infty }(\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right))$$, which can be simplified to$$x =\frac{1}{2}\left(x+\frac{2}{x}\right)$$,which is easily solved to show that $x=\sqrt 2 $ or $ -\sqrt2$, and it is easy to show the limit should be $\sqrt 2 $,since we have already worked out the value of the limit, so also proved the existence of the limit as a side effect.

  1. Does the proof really prove that the sequence has a limit?

  2. Is it possible that we computed out the limit value of a sequence using similar method as above(without proving the existence of the limit before) , but later found that the sequence doesn't have a limit ?

iMath
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  • $x_{n} = ,\sqrt{,2,},\coth\left(,2^{n},\mathrm{arccoth}\left(,{x_{1} \over ,\sqrt{,2,}},\right),\right) \to ,\sqrt{,2,}$ as $n \to \infty$ whenever $x_{1} > 0$. Sorry, I couldn't find the reference but I guess it's somewhere in the DLMF page. – Felix Marin Jun 13 '17 at 23:12
  • @FelixMarin what do you want to hint by giving the sequence and it's limit ? – iMath Jun 16 '17 at 02:56
  • Just to inform that it has a closed form. No more. – Felix Marin Jun 16 '17 at 21:51

6 Answers6

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find the root(s) of $$x^2-2=0$$ by newton raphson method
and choose $\color{red} {x_0>0}$; $$f(x)=x^2-2\to f'(x)=2x\\x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\\x_{n+1}=x_n-\frac{x_n^2-2}{2x_n}$$ $$x_{n+1}=\frac12(2x_n-\frac{x_n^2-2}{x_n})\\ x_{n+1}=\frac12(\frac{x_n^2+2}{x_n})\\ x_{n+1}=\frac12(x_n+\frac{2}{x_n})$$ when your iteration start over a positive number such as $x_0=1$ then $x_n \to +\sqrt2$ It is a method to fond $\sqrt a $ by $$x^2-a=0 \to x=\pm\sqrt a\\(x)=x^2-2\to f'(x)=2x\\x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\\x_{n+1}=x_n-\frac{x_n^2-a}{2x_n}\to \\ x_{n+1}=\frac12(x_n+\frac{a}{x_n})$$

Khosrotash
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  • Thanks for writing so long , but I cannot found any help to my question from your answer – iMath Jun 09 '17 at 14:00
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For $x>0$ we have $x+\frac2x\geq 2\sqrt{x} \sqrt{\frac2x}=2\sqrt2$. So $x_n\ge \sqrt2$. Further $\frac{x_{n+1}}{x_n}=\frac12\left(1+\frac2{x_n^2}\right)\le 1.$ So the sequence converges. Its limit is already known.

Minz
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  • but you have to prove $x_{n}>0$ for almost all $n$ before using the inequality to show $x_n\ge \sqrt2$, any ideas on proving this? – iMath Jun 09 '17 at 09:50
  • @iMath, It's trivial induction: $x_1=1>0$ so $x_2\ge \sqrt2>0\ldots $ – Minz Jun 09 '17 at 09:55
  • Yes! The iteration $x_{n+1}=\frac{1}{2}\left(x_{n}+\frac{2}{x_{n}}\right)$ always gives a positive term provided $x_1$ is positive – iMath Jun 09 '17 at 10:00
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    Particular case of the so-called babylonian method (https://math.stackexchange.com/q/82682). – Jean Marie Jun 09 '17 at 10:14
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  1. No he didn't. He proved that if the limits exists then it is $\sqrt{2}$

  2. However, that is a good first step that you can use by proving afterwards that $y_n=x_n-\sqrt{2}\rightarrow 0$...

Evargalo
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  1. This is not limit on x but limit on n...
user453706
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This only proves that if the sequence has a limit, then the limit must be $\sqrt2$. And even that proof is not totally correct. In fact, the equation has another solution: $-\sqrt2$. But it is easy to see that it cannot be the limit of the sequence.

José Carlos Santos
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Sometimes, but not always, it is much easier to show that if the limit of a sequence or a function exists then it must have a particular value compared to showing that the limit exists. This question is one instance of this phenomenon. Another popular example is $$\lim_{x\to 0}\frac{x-\sin x} {x^{3}}=\frac{1}{6}$$ and the not so popular but very deep prime number theorem $$\lim_{x\to\infty} \frac{\pi(x) \log x} {x} = 1$$ where $\pi(x) $ is the number of primes less than or equal to $x$.

It is better to answer your second question so that you can understand why this technique fails. Consider $a_{n} =n$ so that $a_{2n}=2a_{n}$ which implies $L=2L$ or $L=0$.

Another weird scenario with the above sequence comes from $a_{n+1}=1+a_{n}$ which gives $L=1+L$ which is not possible so that limit $L$ does not exist. You can have a look at a complicated example of a function where proving the existence of limit is not easy, but applying technique of your question is easy.

Paramanand Singh
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