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 : $a_{n-1}&lt;a_{n}$ 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 $\sqrt{6+a_{n}}$
Here's a non-formal trick which can help you on these recurrence problems. But one must be careful and prove that $a_n &gt; 0$ and monotonic first so that $b \geq 0$. You can prove $a_n &gt; 0$ 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 $n \in \mathbb{N}, a_n &lt; 3$. Since we are saying &quot;for all $n$&quot; it must be also true for $n+1 = n'$.
Our base case ($n=1$) is trivial since $a_1 = 
\sqrt{6} &lt; 3$.
Now assuming $p(n):= a_n &lt; 3$ is true, we must show that $p(n+1) = a_{n+1} &lt; 3$ is also true.
In fact,
$$a_n &lt; 3 \Leftrightarrow a_n + 6 &lt; 3 + 6 \Leftrightarrow \sqrt{6 + a_n} &lt; \sqrt{3 + 6} = 3$$
But $\sqrt{6+a_n} = a_{n+1}$ for which $a_{n+1} &lt; 3$ and the condition holds true for all $n$.
So we proved by induction that $0 &lt; a_n &lt; 3$ for all $n \in \mathbb{N}$.
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.