Prove a sequence $(a_n)_{n \geq 1}$ is convergent Given a sequence $(a_n)_{n \geq 1}$ in $\mathbb{R}$, and given that
$$ a_1 > 0 \ \ \ \ \text{and} \ \ \ \ \ a_{n+1} = \sqrt{a_n + 6}\ \ \ (n\geq1)$$
Prove that $(a_n)_{n \geq 1}$ converges.
My thoughts


*

*for $a_1 > 3$, we know that $3<a_{n+1} < a_n \ \ (n\in \mathbb{N})$ (1)

*for $a_1 < 3$, we know that $0<a_{n} < a_{n+1}<3 \ \ (n\in \mathbb{N})$ (2)


Now I now how to prove convergence. However, I'm having difficulty proving my two thoughts about the situation where $a_1 >3$ and $a_1 < 3$. 
Any help on how I can show this correctly?
 A: $$a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{(3-a_n)(2+a_n)}{\sqrt{a_n+6}+a_n}$$
Now, if $a_1<3$ then $a_{n+1}>a_n$, which says that $\{a_n\}$ is increasing and the limit exists. 
If $a_1=3$ then $a_n=3$.
If $a_1>3$ then $\{a_n\}$ is decreasing and the limit exists again.
Let $a$ be our limit.
Thus, $a=\sqrt{6+a}$, which gives $a=3$.
A: For $a_1<3$ let's do this: call $f(x)=\sqrt{x+6}$;
1)$f(x)>0 \forall x>0$ (induction);
2)$f(x+1)>f(x)$ (as you wish, you can also do it by induction);
3)For the monotonical convergence theorem, the function has a limit $l\in \mathbb{R} \cup \{\infty\}$
4) This limit is the fixed point of the function, which is the solution to $x=f(x)$ (in this case is $3$) so the limit is 3.
Specular for the other side.
Try to watch my other similar answer!
A: You have $a_n>0$, for all $n$ (obvious induction).
If $a_n<3$, then $a_{n+1}=\sqrt{a_n+6}<\sqrt{9}=3$. This proves by induction that $a_n<3$, for all $n$, if $0<a_1<3$.
Similarly, for $a_1>3$, prove that $a_n>3$ for all $n$.
You also have
$$
a_{n+1}<a_n\quad\text{if and only if}\quad\sqrt{a_n+6}<a_n
$$
that is $a_n^2-a_n-6>0$, that is $a_n>3$ (taking into account that $a_n>0$).
Therefore the sequence is increasing for $0<a_1<3$ and decreasing for $a_1>3$. In the former case it is bounded above, in the latter it is bounded below. Hence in both cases it converges; if $l$ is its limit, then $l^2-l-6=0$.
A: Here is a more general approach. As Alberto correctly spotted $a_{n+1}=f(a_n)$ where $f(x)=\sqrt{x+6}$. We also have $f'(x)=\frac{1}{2\sqrt{x+6}}$ and $0<f'(x)<1, \forall x\geq0$. Using Mean Value Theorem
$$\left|f(x)-f(y)\right|= f'(c)\left|x-y\right|< k\left|x-y\right|$$
where $0\leq k<1$. Function $f(x)$ has a fixed point at $x=3$, from $f(x)=x \Leftrightarrow x=\sqrt{x+6}$. It's almost perfect condition for Banach fixed point theorem and related theory. I wrote almost because of the constraints imposed by Banach fixed point theorem, specifically domain and range of $f(x)$. But function is ascending, because $f'(x)>0$, and we can always fix a large domain $[0,\alpha]$ to obtain a range $[f(0),f(\alpha)] \subset [0,\alpha]$ (because for larger $\alpha>3$ we have $f(\alpha)<\alpha$) which will also contain the initial point $a_1$.
A: Look at this graph
$$a_n>3\implies 3<\sqrt{a_n+6}<a_n\implies3<a_{n+1}<a_n$$
$$0<a_n<3\implies a_n<\sqrt{a_n+6}<3\implies a_n<a_{n+1}<3$$
