Q:

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:

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.

