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Suppose $a > 0$. Let the sequence $\{x_n\}$ be defined: $x_1=\sqrt a$,
$x_{n+1}=\sqrt{a+x_n}$ for all $n\ge1$. I need to prove that the sequence converges, and I have shown that $x_n<x_{n+1}$, but i am not sure how to prove that there is a bound on the function. I thought it would be $a$ but I think there is a dilemma if $a<1$.

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If $b>0$ is any number such that $b\ge\sqrt{a+b}$, then $b$ is an upper limit: We have $x_1=\sqrt a\le\sqrt{a+b}\le b$ and by induction $x_{n+1}=\sqrt{a+x_n}\le\sqrt{a+b}\le b$. You can use $b=a$ provided $a\ge\sqrt{2a}$, i.e. provided $a\ge2$. But if $a<2$ you can simply take $b=2$ as then $\sqrt{a+2}\le\sqrt 4=2$.

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So this splits a into two cases, a>=2 and a<2. It is possible to find one limit for the sequence, or does the limit need to be broken down? – michelle Apr 26 '13 at 7:03
@michelle As you already showed, the sequence is strictly increasing, hence indeed $x:=\lim_{n\to\infty} x_n$ is also an uppre bound. You probably have determined meanwhile that $x=\frac{1+\sqrt{1+4a}}2$, so of course you may try to prove that $x_n<\frac{1+\sqrt{1+4a}}2$ implies $x_{n+1}<\frac{1+\sqrt{1+4a}}2$. Go ahead, it is possible, but I doubt that it is simpler than the two simple (and motivated) cases in my answer. But you may also be lucky with trying $b=a+2$ for all $a$, try it (and check if the proof really becomes simpler). – Hagen von Eitzen Apr 26 '13 at 14:34

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