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$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$

i am trying to calculate the limit of $a_n:=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+}}}}..$ with $a_0:=1$ and $a_{n+1}:=\sqrt{1+a_n}$ i am badly stuck not knowing how to find the limit of this sequence and where to start the proof. i did some calculations but still cannot figure out the formal way of finding the limit of this sequence. what i tried is:
$$(1+(1+(1+..)^\frac{1}{2})^\frac{1}{2})^\frac{1}{2}$$ but i am totally stuck here

Community
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doniyor
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  • The limit is the positive root of $l^2-l-1=0$. – user 1591719 Dec 30 '12 at 09:16
  • how do your come to this? can you pls explain a bit? – doniyor Dec 30 '12 at 09:20
  • I think this is a duplicate. I'm trying to find that question. There you may find all needed explanations. – user 1591719 Dec 30 '12 at 09:23
  • now i got i think. you mean that assuming $a$ be a limit, then $a^2=1+a$. so the root of this equation is the limit of the sequence? – doniyor Dec 30 '12 at 09:24
  • I've had a problem understanding this method a few weeks ago, too, this question might help you understand. – Dahn Dec 30 '12 at 09:31
  • @DahnJahn, thanks Dahn, perfect – doniyor Dec 30 '12 at 09:32
  • i getting $x_1=\frac{1+\sqrt{5}}{2}$ and $x_2=\frac{1-\sqrt{5}}{2}$, so are those limits? if the sequence converges, then there is only ONE limit, why i am getting 2 here? – doniyor Dec 30 '12 at 09:38
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    You get two solutions because there are two possible signs for a square root and you have both because you squared the equation before solving it - which means that the algebra cannot distinguish between them. – Mark Bennet Dec 30 '12 at 09:49
  • @Mark, thanks, beautiful sentence! – doniyor Dec 30 '12 at 09:50
  • You can prove that the limit exists by showing that your $a_n$ is an increasing sequence, and also that it is bounded above by 2. – Mark Bennet Dec 30 '12 at 09:51
  • @MarkBennet that was indeed beautifully put. I thought I understand the process completely, but this gave me a new insight into mathematics, thanks! – Dahn Dec 30 '12 at 09:52
  • @Dahn, the same to me. – doniyor Dec 30 '12 at 09:57
  • I find disturbing that the accepted answer, even including the discussion in its comments, fails to provide a complete solution without mentioning it. – Did Dec 30 '12 at 10:39
  • i took the torch on the way because of this equation, because i didnot need further guidance, then i accepted the answer – doniyor Dec 30 '12 at 10:49

2 Answers2

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We (inductively) show following properties for sequence given by $a_{n+1} = \sqrt{1 + a_n}, a_0 =1$

  1. $a_n \ge 0$ for all $n\in \Bbb N$
  2. $(a_n)$ is monotonically increasing
  3. $(a_n)$ is bounded above by $2$

Then by Monotone Convergence Theorem, the sequence converges hence the limit of sequence exists. Let $\lim a_{n} = a$ then $\lim a_{n+1} = a$ as well. Using Algebraic Limit Theorem, we get

$$ \lim a_{n+1} = \sqrt{1 + \lim a_n} \implies a = \sqrt {1 + a} $$

Solving above equation gives out limit. Also we note that from Order Limit Theorem, we get $a_n \ge 0 \implies \lim a_n \ge 0$.

S L
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Hint: First of all show that the sequence conveges. Then if $a_n\to L$ when $n\to \infty$ assume $L=\sqrt{1+L}$ and find $L$.

Mikasa
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    Just as an additional hint: Showing that the sequence converges would be usually done by showing that it is monotone and bounded (and therefore converges to a number within that bound). – Dahn Dec 30 '12 at 09:37
  • i getting $x_1=\frac{1+\sqrt{5}}{2}$ and $x_2=\frac{1-\sqrt{5}}{2}$, so are those limits? if the sequence converges, then there is only ONE limit, why i am getting 2 here? – doniyor Dec 30 '12 at 09:38
  • @DahnJahn: Yes the OP should show what you noted him first before doing any handy manipulation on $a_n$. Thanks for noting me and him. ;-) – Mikasa Dec 30 '12 at 09:40
  • You need to show (using induction) that the sequence is bounded from below by 1 (if $a_n\geq 1$ then $a_{n+1}\geq 1$). edit - in fact, 0 would do, too. – Dahn Dec 30 '12 at 09:41
  • @doniyor: We see that all terms in $a_n$ are positive. – Mikasa Dec 30 '12 at 09:42
  • @BabakSorouh, oh okay you are right – doniyor Dec 30 '12 at 09:46