Limit of a sequence given by $x_n = \sqrt{3 + x_{n-1}}$ as n approaches infinity

Question

Let $x_1 = 1$ and for $n \ge2, x_n = \sqrt{3 + x_{n-1}}$, I just want to know how to determine the limit.

This post seems related: http://math.stackexchange.com/questions/115501/sqrtc-sqrtc-sqrtc-cdots-or-the-limit-of-the-sequence-x-n1-sq And you can find several related questions among the posts linked there. — Martin Sleziak, Oct 28 '15 at 14:48
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score 3 · Answer 1 · answered Oct 28 '15 at 13:27

3

Hint: prove that your sequence is increasing and bounded, hence converging to its supremum, then notice that there is only one real number $r$ such that $r=\sqrt{r+3}$.

answered Oct 28 '15 at 13:27

Jack D'Aurizio

353,855

score 1 · Answer 2 · answered Oct 28 '15 at 14:08

1

Hint: If you know the Banach fixed-point theorem, then $f(x)=\sqrt{3+x}$ has Lipschitz constant $1/4$ on $[1,\infty)$.

answered Oct 28 '15 at 14:08

Lutz Lehmann

126,666

Limit of a sequence given by $x_n = \sqrt{3 + x_{n-1}}$ as n approaches infinity

2 Answers2