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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

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marked as duplicate by Rahul, Did, Davide Giraudo, Fabian, QiL Dec 30 '12 at 14:13

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.

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

2 Answers 2

up vote 7 down vote accepted

let $a =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+}}}}...$ be it's limit.

Squaring we get $a^2 = 1 + a$

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can you please expand your "Squaring we get $a^2=1+a$" so that i can see your steps in between –  doniyor Dec 30 '12 at 9:17
Assume that $a$ be it's limit, then inside the square root, you get the same pattern since there are infinite no of terms. –  Santosh Linkha Dec 30 '12 at 9:21
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? –  doniyor Dec 30 '12 at 9:26
@doniyor: Yes, it is what he meant. :-) –  Babak S. Dec 30 '12 at 9:28
oh okay, then i have to solve the equation $a^2-a-1=0$ –  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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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 Jahn Dec 30 '12 at 9:37
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 9:38
@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. ;-) –  Babak S. Dec 30 '12 at 9:40
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 Jahn Dec 30 '12 at 9:41
@doniyor: We see that all terms in $a_n$ are positive. –  Babak S. Dec 30 '12 at 9:42

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