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Possible Duplicate:
$\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, 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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    i took the torch on the way because of this equation, because i didnot need further guidance, then i accepted the answer2012-12-30

2 Answers 2

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

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    ... my please :)2019-04-25
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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 $L$.

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    @BabakSorouh, oh okay you are right2012-12-30