Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.