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So, I have been trying to solve this problem since last night and finally now decided to have some help , here. If -

$ y_1= \sqrt p$, $p> 0$, and $y_{n+1} = \sqrt{p+y_{n}}$ for all $n \in \mathbb{N}$. I wish to show that $y(n)$ converges and find its limit.

It requires use of monotone convergence theorem so we will have to prove that it is bounded and monotone, first.
I solved a question just like this one one I was given some real value of $p$ but with some unknown like $p$ here I don't know how to start with proving it to be bounded or monotone. I f somebody can just help me with that I would be truly grateful.

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PROTIP: These forums are a great place to practice your LaTeX! – Simon Hayward Nov 16 '12 at 11:10
@SimonHayward thanks for editing. – shrey Nov 16 '12 at 11:36
up vote 2 down vote accepted

Call $z$ the unique fixed point of the function $u:x\mapsto\sqrt{p+x}$, hence $z=\frac12(1+\sqrt{1+4p})$.

For every $n\geqslant1$, $y(n+1)=u(y(n))$, hence the result you are after follows from the three points below:

  1. $y(1)\lt z$.
  2. If $y(n)\lt z$, then $y(n+1)\lt z$.
  3. If $y(n)\lt z$, then $y(n+1)\gt y(n)$.

If one of these is a problem, mention it.

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Since $y_1=\sqrt{p}$ and $p \gt 0$. One way of looking at it is by induction: $y_2=\sqrt{y_1+p}\gt y_1$. Then assume that is is true for $n$(i.e $y_n \gt y_{n-1}$). Then $y_{n+1}=\sqrt{y_n+p}\gt\sqrt{y_{n-1}+p}=y_n$. Hence the sequence ${y_n}$ is increasing.

The other way is to let $f(x)=\sqrt{p+x}$. Then $f'(x)=\dfrac{1}{2\sqrt{p+x}}$ which is always greater than $0$. hence $f(x)$ is increasing. Now as you know that $y_2 \gt y_1$, $f(y_2) \gt f(y_1)\implies y_3 \gt y_2$

Going on like this will give you the required conclusion. Rest it is easy to see that the sequence is bounded. Hence it converges.

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