Limit of sequence $a_n = \sqrt{r + a_{n-1}}$

Question

Problem says to let $r>0$ and $a_0=\sqrt{r}$ and $a_n = \sqrt{r + a_{n-1}}$ for $n\ge1$

I know that $a_n$ is such that $a_0 < a_n < \sqrt{r} +1$ by showing

$a_1 = \sqrt{r+a_0} = \sqrt{r+\sqrt{r}} < \sqrt{r} +1$ and then using induction. But I'm not sure where to go from there.

sorry I messed up the Latex in the title – Corran Horn Oct 10 '17 at 01:21 — Corran Horn, Oct 10 '17 at 01:21
You are able to edit your question (and its title)... – Eric Towers Oct 10 '17 at 01:25 — Eric Towers, Oct 10 '17 at 01:25
thanks for the edits, I'm new to this (obviously) – Corran Horn Oct 10 '17 at 01:32 — Corran Horn, Oct 10 '17 at 01:32

score 0 · Accepted Answer · answered Oct 10 '17 at 05:10

We can assume that a solution will be invariant under further iteration, and thus: $$a_{f}=\sqrt{r+a_{f}}\\\to a_f^2=r+a_f\\\to a_f = \frac{1\pm\sqrt{1+4r}}{2}$$

And given that we know our answer should be positive:

$$a_f=\frac12(1+\sqrt{1+4r})$$

($a_f$ is short for $a_{final}$, the value the sequence converges to.)

Limit of sequence $a_n = \sqrt{r + a_{n-1}}$

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