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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

  • $\begingroup$ The limit is the positive root of $l^2-l-1=0$. $\endgroup$ – user 1591719 Dec 30 '12 at 9:16
  • $\begingroup$ how do your come to this? can you pls explain a bit? $\endgroup$ – doniyor Dec 30 '12 at 9:20
  • $\begingroup$ I think this is a duplicate. I'm trying to find that question. There you may find all needed explanations. $\endgroup$ – user 1591719 Dec 30 '12 at 9:23
  • $\begingroup$ 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? $\endgroup$ – doniyor Dec 30 '12 at 9:24
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    $\begingroup$ 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. $\endgroup$ – Mark Bennet Dec 30 '12 at 9:49

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 $a_n \ge 0 \implies \lim a_n \ge 0$.

  • $\begingroup$ can you please expand your "Squaring we get $a^2=1+a$" so that i can see your steps in between $\endgroup$ – doniyor Dec 30 '12 at 9:17
  • $\begingroup$ Assume that $a$ be it's limit, then inside the square root, you get the same pattern since there are infinite no of terms. $\endgroup$ – Santosh Linkha Dec 30 '12 at 9:21
  • $\begingroup$ if $a_n$ converges to $a$, then $a_{n+1}$ also converges to $a$,right? if yes, then $a_{n+1}=\sqrt{1+a_n}$ converges to $a=\sqrt{1+a}$, is this what you mean? $\endgroup$ – doniyor Dec 30 '12 at 9:26
  • $\begingroup$ @doniyor: Yes, it is what he meant. :-) $\endgroup$ – Mikasa Dec 30 '12 at 9:28
  • $\begingroup$ oh okay, then i have to solve the equation $a^2-a-1=0$ $\endgroup$ – doniyor Dec 30 '12 at 9:30

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 $L$.

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    $\begingroup$ 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). $\endgroup$ – Dahn Dec 30 '12 at 9:37
  • $\begingroup$ 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? $\endgroup$ – doniyor Dec 30 '12 at 9:38
  • $\begingroup$ @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. ;-) $\endgroup$ – Mikasa Dec 30 '12 at 9:40
  • $\begingroup$ 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. $\endgroup$ – Dahn Dec 30 '12 at 9:41
  • $\begingroup$ @doniyor: We see that all terms in $a_n$ are positive. $\endgroup$ – Mikasa Dec 30 '12 at 9:42

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