3
$\begingroup$

I'm trying to prove that this sequence $(x_n)$, where $x_1 =\sqrt a$ and $x_{n+1}=\sqrt{a +x_n}$ has a limit, then I would like to find the limit of

$L=\sqrt {a+\sqrt{a+\sqrt{a+\sqrt{a...}}}}$

It's easy to find this limit and prove that this sequence is monotone (induction over $\mathbb N$). What I found difficult is prove that this sequence is bounded.

I need help in this part.

Thanks a lot.

$\endgroup$

marked as duplicate by MJD, njguliyev, Henry T. Horton, Thomas Andrews, Dan Rust Sep 5 '13 at 16:05

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.

3
$\begingroup$

When $a\ge1$, we have $x_1=\sqrt{a}\le a<2a$ and by mathematical induction, $$x_{n+1}^2 = a + x_n < a+2a=3a < 4a^2\ \Rightarrow\ x_{n+1}<2a.$$ When $a<1$, the sequence is dominated by the analogous sequence with $a=1$ and hence it is bounded.

$\endgroup$
  • $\begingroup$ I didn't understand, when a=2, we have $\sqrt {2+\sqrt{2+\sqrt{2+\sqrt{2…}}}}$, it isn't a constant sequence. $\endgroup$ – user42912 Feb 19 '13 at 10:36
  • $\begingroup$ @user42912 Oops, I misread the question. Thanks for pointing out the error. Will delete this answer in a couple of minutes. $\endgroup$ – user1551 Feb 19 '13 at 10:37
  • $\begingroup$ no problem, thank you for trying to help me :) $\endgroup$ – user42912 Feb 19 '13 at 10:38
  • $\begingroup$ @user42912 I've tried to fix the proof. Please see if it's OK now. $\endgroup$ – user1551 Feb 19 '13 at 10:47
  • $\begingroup$ it seems ok! thanks again. $\endgroup$ – user42912 Feb 19 '13 at 10:54
1
$\begingroup$

If the limit exists, it is obvious to see that $$ \lim_{n\rightarrow \infty} a_{n+1} = \lim_{n \rightarrow \infty} a_n$$ So lets call the limit $x$ than $$x=\sqrt{a+x}$$ For the bound, i would use banach fix point theorem. The fix point theorem together with the fact, that the root is a strict contraction. So we know $$x_1\leq \sqrt{a} + \sqrt{\sqrt{a}}$$ For $a$ sufficiently large $a$ will be a upper bound and $0$ a lower bound. For $a$ not sufficiently large we don't need a proof, since we know it is lower bounded by $0$ and monotone increasing, so we can take the upper bound of the sufficiently large $a$ (monotonicity of the root).

$\endgroup$
  • $\begingroup$ How would you use this theorem? I couldn't figure out how to apply in this question. $\endgroup$ – user42912 Feb 19 '13 at 10:18
  • $\begingroup$ the root is a strict contraction $\endgroup$ – Dominic Michaelis Feb 19 '13 at 10:25
  • $\begingroup$ I didn't understand, please add a little bit more information in your answer. Thank you very much $\endgroup$ – user42912 Feb 19 '13 at 10:32
  • $\begingroup$ better now ? sry if it was to short $\endgroup$ – Dominic Michaelis Feb 19 '13 at 10:50
  • $\begingroup$ sorry but I didn't understand yet. Maybe I don't have the maturity to understand your answer yet. Anyway, thank your for trying to help me. $\endgroup$ – user42912 Feb 19 '13 at 11:00

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