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$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, 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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marked as duplicate by Rahul, Babak S., Martin Sleziak, Hans Lundmark, Did Jan 4 '13 at 12:54

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The first series converges to 2, but that doesn't mean it is equal to 2. – Ram Jan 4 '13 at 11:41
@Ram What do you mean by that? – EuYu Jan 4 '13 at 11:48
Duplicate of – Rahul Jan 4 '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 4 '13 at 11:54
@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 4 '13 at 12:05
up vote 3 down vote accepted

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.

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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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Nice link! Helpful answer! +1 – amWhy Feb 24 '13 at 0:08

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