Finding limit point of recursively defined sequence $a_{n+1}=\sqrt{2+a_n}$.

Question

Problem: $\{a_n\}_{n\in \mathbb N}, \quad a_{n+1}=\sqrt{2+a_n}, \quad \forall n\geq 1, \quad a_1=\sqrt{2}$

Solution:

We assume $a_n$ converges. Then is $\lim_{n\to\infty}a_n=\lim_{n\to\infty}a_{n+1}$

So we get

$$$a_n=\sqrt{2+a_n} \Rightarrow a_n^2-a_n-2=0$$

which is solved by $a_n=-1$ and $a_n=2$ but since $a_n\geq 0 \forall n$ we only find 2 to be a limit point.

Question: Is that argument okay? I'm not sure, the main part I don't like about is, is the assumption of convergence. Does it work?

I know there are differet solutions, but how good is the one above? Is it valid?

Answer 1

If $(a_n)$ is convergent, then denote the limit of the sequence by $a$. Then we get

$$a=\sqrt{2+a}.$$

Hence $a=-1$ or $a=2$ (and not $a_n=-1$ or $a_n=2$ as you wrote). Since all $a_n \ge 0$, we see that $a_n \to 2$.

All this considerations are made under the assumption that $(a_n)$ is convergent !

Thus it remains to show that $(a_n)$ is convergent .

To this end show that $(a_n)$ is increasing and bounded.

Answer 2

We need to refer to monotonic sequences theorem and show by induction that $a_n$ is strictly increasing and bounded above, then we can claim that the limit exists and therefore set

$$L=\sqrt{2+L}\implies L=2$$

Answer 3

$a_n <2$ implies $a_{n+1} =\sqrt {2+a_n} < \sqrt {2+2}=2$. Since $a_1 <\sqrt 2$ it follows by induction that $a_n <\sqrt 2$ for all $n$. Next $a_{n+2}^{2}- a_{n+1}^{2}=(2+a_{n+1})-(2+a_n)= a_{n+1}-a_n$. This gives $(a_{n+2}-a_{n+1})(a_{n+2}+a_{n+1})=(a_{n+1}-a_n)$. Since $a_{n+2}+a_{n+1} >0$, we can conclude that $a_{n+1}-a_n$ has the same sign for all $n$. This makes $\{a_n\}$ a monotonic bounded sequence, so it has a limit. As you have already shown the limit must be $2$.