How to prove a sequence is strictly increasing and bounded above? For this proof, I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Let $\{a_n\}$ be a sequence defined recursively by
$a_1 = \sqrt{6}$
$a_{n+1} = \sqrt{6+5a_n}, n = 1,2,3,...$
Prove that $\{a_n\}$ is strictly increasing and strictly bounded above
$a_2 = \sqrt{6+5\sqrt{6}}< \sqrt{6+5*15} = 9$
$a_3 = \sqrt{6+5a_2}$
$a_3 = \sqrt{6+5\sqrt{6+5 \sqrt{6}}}$
wts there exists an M st $a_n \le M$
Choose $M = 9$
Base Case: $a_1 = \sqrt{6}<\sqrt{81} = 9$
Induction step:
Let k in the natural numbers be arbitrary
Assume $a_k \le 9$
$a_{k+1} = \sqrt{6+5a_k} < \sqrt{6+75} = \sqrt{81} = 9$
Therefore by induction, $a_n \le 9$
$a_{n+1} = \sqrt{6+5a_n}$
$a_{n+2} = \sqrt{6+5a_{n+1}}$
$a_{n+1} \le a_{n+2}$
Therefore, the sequence is strictly increasing and strictly bounded above

It is correct globally, but I think you are not expressing yourself clearly about why the sequence is strictly increasing. The base case is $a_1<a_2$, which is clearly true. Andn, if $a_n<a_{n+1}$, then $6+5a_n<6+5a_{n+1}$ and therefore $\sqrt{6+5a_n}<\sqrt{6+5a_{n+1}}$. But this means that $a_{n+1}<a_{n+2}$.
