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Possible Duplicate:
$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

Some time ago I was playing with a calculator and I found the following relation $$2 = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}}$$ In fact I found more, I found that $$r = \sqrt{r(r - 1) + \sqrt{r(r - 1) + \sqrt{r(r - 1) + \sqrt{r(r - 1) + \cdots}}}}$$ if $r > 1$, but I couldn't give a formal proof and I still can't.

Note: If you solve $r(r - 1) = 1$ then you'll find an interesting property of the golden number.

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Diego Silvera
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  • The first series converges to 2, but that doesn't mean it is equal to 2. – Ram Jan 04 '13 at 11:41
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    @Ram What do you mean by that? – EuYu Jan 04 '13 at 11:48
  • Duplicate of http://math.stackexchange.com/q/115501/856 –  Jan 04 '13 at 11:48
  • @Eu Yu, I am saying value is different from limit. The correct sentence is, $ \sqrt{2 + \sqrt {2 + \sqrt {2 ....}}}$ converges to $ 2$, you can't say that $ \sqrt{2 + \sqrt {2 + \sqrt {2 ....}}}$ is equal to $2$ which is wrong, since $ 2\neq \sqrt{2 + \sqrt {2 + \sqrt {2 ....}}}$ – Ram Jan 04 '13 at 11:54
  • I think the form of this question adds something to the duplicate - it illuminates the structure and gives the direct equation for the limit $L(L-1)=r(r-1)$. – Mark Bennet Jan 04 '13 at 11:58
  • @Ram, you are giving bad advice in my opinion. If you really want to be correct about it, you have to say that the sequence $(x_n)$ converges to 2, where $x_n=\sqrt{2+\sqrt{2+\sqrt{2+\ldots+\sqrt 2}}}$ ($n$ times). Your version is just as inaccurate as the OP's version (not that it matters $-$ the OP's version is perfectly acceptable in a maths forum). – TonyK Jan 04 '13 at 12:02
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    @Ram It is meaningless to talk about the value of the nested radical unless there is convergence. In the case that the sequence of nested radicals $$\left{\sqrt{2},\ \sqrt{2+\sqrt{2}},\ \cdots\right}$$ converges then we define the value of the nested radical to be equal to the limit of the sequence. This is the same approach taken with infinite series. If the limit is in fact equal to $2$, then there is absolutely nothing wrong with the equality. – EuYu Jan 04 '13 at 12:05
  • @Eu Yu, yes agreed, but the post doesn't say anything about convergence or limit. – Ram Jan 04 '13 at 12:09
  • @Ram: And your comment doesn't say anything about what you mean by $\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$. Which is a much worse crime, given that you are taking it upon yourself to correct the OP. If you had taken the trouble to define it, you would have seen that your final assertion is simply false: $\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}$ is, in fact, equal to 2. – TonyK Jan 04 '13 at 18:49
  • @TonyK, thanks for your comment, and regarding my last assertion, give me some time to correct myself, I mean I can't give a formal proof for that, so I will learn that. – Ram Jan 05 '13 at 05:39

2 Answers2

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As @Ram commented, if you set $a_{n+1}= \sqrt{2 + a_n}$, provided you show $a_n$ converges so there is an $L$ such that $a_n\to L$ when $n\to\infty$ and so you can solve the following equation to find $L$: $$L=\sqrt{2+L}$$. The general link is $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

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Mikasa
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Let $x_1=\sqrt{r(r-1)}$ and $x_{n+1}=\sqrt{r(r-1)+x_n}$ for $n \geq 1$

Then you can use $0<x_n<r$ to show that the $x_n$ are bounded above and increasing(*), which means they tend to a limit. Let that limit be x, then we have:$$x=\sqrt{r(r-1)+x}$$ which can be squared to give the solution $x=r$ (and the inadmissible $x=1-r$).

For (*) we have $$x_{n+1}-x_n=\sqrt{r(r-1)+x_n}-x_n$$ and we want to show that this is positive, so we multiply the rhs by the positive number $\sqrt{r(r-1)+x_n}+x_n$ to obtain:$$r(r-1)+x_n-x_n^2=r(r-1)-x_n(x_n-1)$$ and it remains to show that this is positive for the values required.

Mark Bennet
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