Define sequence $s_1 = \sqrt{2}$ and $s_{n+1} = \sqrt{2+s_n}$. Find $\lim_{n \rightarrow \infty} s_n$ and prove this convergence.

Question

I'm completely at a lost as to how to solve this. A few things I do know is that $s_n > -2$ because square root can't take negative values. So this implies that $s_n$ is bounded by -2? I also know of a theorem that says that $s_n$ will converge if it is a monotone bounded sequence. The sequence is also increasing. So by the theorem, it will converge to $\lim_{n \rightarrow \infty} s_n = \sup \{ s_n : n \in \mathbb{N}\}$?

I'm not sure if this is correct and I'm also struggling with trying to prove this. Help would be appreciated.

Have you tried simulating it? Easy to set up on a machine. Should give you a good sense of how this performs. Also: if the limit, $L$, exists then $L=\sqrt {2+L}$. — lulu, Nov 01 '16 at 20:15
Side note: a few trials should dispel the notion that "the sequence is also increasing". — lulu, Nov 01 '16 at 20:18
"I also know of a theorem that says that sn will converge if it is a monotone bounded sequence" Excellent, thus, is $(s_n)$ a monotone bounded sequence? — Did, Nov 01 '16 at 20:21
Special case of $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ — YuiTo Cheng, Jul 11 '19 at 08:13

score 0 · Answer 1 · answered Nov 01 '16 at 20:21

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You should consider two cases. One when $t \ge 4$ and the other when $t \le 4$.

For the first one using inductive reasoning we have that:

$$s_{n+1} \le s_n \iff \sqrt{2 + s_n} \le s_n \iff 2 + s_n \le s_n^2 \iff s_n^2 - s_n - 2 \ge 0$$ $$ \iff (s_n - 2)(s_n + 1) \ge 0$$

which is true as $s_n \ge 2$.

Also we can prove that:

$$s_{n+1} \ge 2 \iff \sqrt{2 + s_n} \ge 2 \iff 2 + s_n \ge 4 \iff s_n \ge 2$$

Therefore $s_n$ is decreasing sequence and bounded from above, so it's convergent.

If $t\le 4$, then prove that $s_n \le 2$ and $s_n$ is an increasing sequence.

To find the limit set $\lim_{n \to \infty} s_n = L$ and take the limit of the recursive relation to get:

$$L = \sqrt{2+L} \iff L^2 = 2 + L \iff (L-2)(L+1) = 0$$

Obviously we must have $L=2$, so $\lim_{n \to \infty} s_n = 2$

answered Nov 01 '16 at 20:21

Stefan4024

35,843

Thanks for the reply! I had a typo on the title. It should read $s_1 = \sqrt{2}$. What would change in this case? I'm guessing we would have only the case where $s_n > -2$? – Nikitau Nov 02 '16 at 04:41
@Nikitau Note that $s_n$ can't be negative. Now do the same, instead of $t\ge 4$ and $t<4$ check the cases when $t \ge 2$ and $t <2$ – Stefan4024 Nov 02 '16 at 05:20
Quick follow-up question, is there way to show the entire seq is bounded above by 2? I'm not sure how to justify the cases – Nikitau Nov 02 '16 at 23:23
@Nikitau It depends. If $s_1<\sqrt{2}$ then the sequence is bounded by $\sqrt{2}$ from above, otherwise it's bounded by the same bound, but from below. – Stefan4024 Nov 02 '16 at 23:25
Oh I get it. So in the first line of where you show $s_{n+1} \leq s_n$, we're showing that it is a decreasing sequence. The second line shows that $s_n$ is indeed bounded. – Nikitau Nov 03 '16 at 01:50

Define sequence $s_1 = \sqrt{2}$ and $s_{n+1} = \sqrt{2+s_n}$. Find $\lim_{n \rightarrow \infty} s_n$ and prove this convergence.

1 Answers1