# 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

• I don't understand your calculation. You appear to assert that $a_2=5\sqrt 6$ but this is not true. – lulu Mar 31 '18 at 18:34
• its root 6 plus five times root 6 – dg123 Mar 31 '18 at 18:49
• Your formula for $a_3$ incorrectly replaces $a_2$ with $5\sqrt 6$. – lulu Mar 31 '18 at 18:50
• Check the edit I made. – dg123 Mar 31 '18 at 18:54
• I changed it, take a look – dg123 Mar 31 '18 at 18:56

It is correct globally, but I don't 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 $a_{n+1}<a_{n+2}$.
• @dg123 First: you did not mention that $a_1<a_2$. Second: you didn't justify the equality $a_{n+1}\leqslant a_{n+2}$. – José Carlos Santos Mar 31 '18 at 18:56