'Continued roots''

Question

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.

Answer 1

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.

Answer 2

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 $x_{n+1} = \sqrt{c+x_n}$