how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded? I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$
 A: Here's a non-formal trick which can help you on these recurrence problems. But one must be careful and prove that $a_n > 0$ and monotonic first so that $b \geq 0$. You can prove $a_n > 0$ easily with induction and since you've already proven it is increasing let's get straight to the point.
$$\lim a_{n+1} = \lim \sqrt{6+a_n} = b = \lim a_n$$
So $b = \sqrt{6+b} \Leftrightarrow b^2-b-6 = 0 \Leftrightarrow (x+2)(x-3) = 0 \Leftrightarrow b = 3 \lor b = -2$. Since $b \geq 0$ then $b = 3$.
We conclude that the limit of this monotonic and positive recurrence sequence is $3$, so it becomes closer and closer to 3 from below. Meaning $a_n \rightarrow 3^-$.
This should not be your proof; just a side calculation that helps you get the supremum for $\{a_n\}_{n=1}^{\infty}$. We will use it to show that $a_n$ is bounded above.
Now we must demonstrate it with a formal proof (by induction) that for all $n \in \mathbb{N}, a_n < 3$. Since we are saying "for all $n$" it must be also true for $n+1 = n'$.
Our base case ($n=1$) is trivial since $a_1 = 
\sqrt{6} < 3$.
Now assuming $p(n):= a_n < 3$ is true, we must show that $p(n+1) = a_{n+1} < 3$ is also true.
In fact,
$$a_n < 3 \Leftrightarrow a_n + 6 < 3 + 6 \Leftrightarrow \sqrt{6 + a_n} < \sqrt{3 + 6} = 3$$
But $\sqrt{6+a_n} = a_{n+1}$ for which $a_{n+1} < 3$ and the condition holds true for all $n$.
So we proved by induction that $0 < a_n < 3$ for all $n \in \mathbb{N}$.
A: $a_n\leq 3$ can be proved by induction:
Induction base: $a_1=0\leq 3$.
Assume that $a_n\leq 3$, then we have $a_{n+1}=\sqrt{6+a_n}\leq \sqrt{6+3}=3$
A: Claim. If $a_n$ is increasing, then $a_n<3$.
Otherwise, if $a_n\ge 3$, for some $n\in\mathbb N$, then
Case A. $a_n=3$, then $a_{n+1}=\sqrt{a_n+6}=\sqrt{3+6}=\sqrt{9}=3$. Hence $a_n$ is not increasing.
Case B. $a_n>3$. Then $a_{n+1}>a_n>3$ and
$$
a_{n+1}-a_n=\sqrt{a_n+6}-a_n=\frac{\sqrt{(a_n+6}-a_n)(\sqrt{a_n+6}-a_n)}{\sqrt{a_n+6}+a_n}\\=\frac{a_n+6 -a_n^2}{\sqrt{a_n+6}+a_n}=-\frac{(a_n+2)(a_n-3)}{\sqrt{a_n+6}+a_n}<0.
$$
Contradiction.
Hence, $a_n$ is upper bounded by $3$ and lower bounded by $a_1$.
A: Hint: Let $f(x)=\sqrt{6+x}$ for $x \ge -6$. Prove that

*

*$f(x) \ge x$ when $ -6 \le x \le 3$


*$f(x) \le x$ when $ 3 \le x$
Apply this result to $a_{n+1}=f(a_n)$ and conclude that the sequence is monotone.
