Prove convergence of sequence defined recursively $a_{n+1} = \sqrt{6+a_n}$ We have a sequence:
$$a_1=\sqrt{6}$$
$$a_{n+1} = \sqrt{6+a_n}$$
The problem is to check convergence and then find the limit.
We know that sequence converges when it is monotonic and bounded. With a little trying I can guess, that this sequence must be bounded by 3 and it's monotonic. Then I can prove my guess:
Proving upper boundary I guessed $a_n<3$:


*

*For $a_1 = \sqrt{6} < 3$

*$a_{n+1}= \sqrt{6+a_n}$

*$a_{n+1} = \sqrt{6+3} = 3$


However, I personally don't like this proof. It requires guessing and checking if I was right.
Later I could use this and prove monotonicity, but I'd like to ask you for some better proofs you may know.
 A: You just need to guess that it is indeed bounded. In that case, the limit if it exists will be a bound (since the sequence is monotonic). In this case, it is easy to see that the limit if it exists is 3 (by solving $a^2 - a - 6 = 0$). So $a_n < 3$ does indeed hold, it is just a matter of proving that by induction.
A: To avoid "guessing" first establish the fixed points of the recurrence, i.e. the solutions of $$a=\sqrt{6+a}$$ or $$a^2-a-6=0$$ This gives $a=3, a=-2$ and since $a$ is positive the likely limit is $3$ (it could be infinite, for example).
Note that $a_1=\sqrt 6\lt 3$ and if $a_n\lt 3$ then $$a_{n+1}^2=a_n+6\lt 9 \dots (1)$$so $a_{n+1}\lt 3$ and the elements of the recurrence are positive and bounded above by $3$.
Now note also that $a_{n+1}^2-a_n^2=a_n-a_{n-1}$ (by subtracting successive instances of the equation which appears in $(1)$ above) which shows that the recurrence is monotonic. It is easy to show that it is increasing.
A: Define the function $f$ on $\mathbb{R}_+$ by
$$\forall x\in\mathbb{R}_+,\ f(x)=\sqrt{6+x}.$$
For all $x\in\mathbb{R}_+$,
$$f'(x)=\frac1{2\sqrt{6+x}}$$
hence
$$\bigl\lvert f'(x)\bigr\rvert\leq\frac1{2\sqrt6}<1.$$
Hence $f$ is a contraction mapping on $\mathbb{R}_+$ and since the codomain of $f$ is included in the domain of $f$ and $\mathbb{R}_+$ is closed we can conclude, by the fixed point theorem, that $f$ possesses a unique fixed point $c\in\mathbb{R}_+$; moreover, for all $a_0\in\mathbb{R}_+$, the sequence $(a_n)_{n\in\mathbb{N}}$ defined by
$$\forall n\in\mathbb{N},\ a_{n+1}=f(a_n)$$
converges to $c$. Moreover,
$$\forall n\in\mathbb{N},\ \lvert a_n-c\rvert\leq\frac{k^n}{1-k}\lvert a_0-a_1\rvert$$
where
$$k=\frac1{2\sqrt6}.$$
Conclusion: your sequence converges to $c$ and you have a rough idea of how it converges to $c$.
It's quite easy to determine $c$ explicitly by solving $f(c)=c$ (left to the reader—but you already have it: $c=3$).
A note about the variations of $(a_n)$: since $f$ is increasing, $(a_n)$ is monotone, so you only need to compare $a_0$ and $a_1$. In your case, $a_0=\sqrt{6}$ and $a_1=\sqrt{6+\sqrt{6}}>a_0$, hence $(a_n)$ is increasing.
That was a good proof I know.
A: Once you guessed that the limit, if it exists, is $3$ because this is the intercept of the first diagonal $y=x$ and of the graph $y=\sqrt{x+6}$, you can check the convergence, using the identity
$$
a_{n+1}-3=\frac{a_n-3}{a_{n+1}+3}.
$$
Since $a_{n+1}\geqslant0$, this implies that
$$
|a_{n+1}-3|\leqslant\frac13\cdot|a_n-3|.$$
