calculus - Proof by induction for recursive sequence with no explicit formula.

The problem I am trying to solve is: "show that the sequence defined by $a_1=6$ and $a_{n+1}=\sqrt{6+a_n}$ for $n\ge 1$ is convergent, and find the limit."

So I know that I need to use proof by induction to show that the sequence is decreasing, and then show that it has a greatest lower bound of $3$. And then by the Monotone convergence theorem I know it converges to $3$.

I tried to find an explicit formula for the sequence but I was unsuccessful. So my problem is that I don't know how to use induction on a non-explicit defined recursive sequence.

Answer

First we prove by induction that $a_n>3$. It's true for $n=1$. Assuming $a_n>3$, we know $a_n+6>3^2$ so $\sqrt{a_n+6}>3$ or $a_{n+1}>3$. Thus we established the lower bound $3$.

Now we see that $x^2-x-6$ is a strictly increasing polynomial for $x>3$, and has a root at $x=3$, thus, $x^2-x-6>0$ for $x>3$: We see that $a_n^2-a_n-6>0$, which we can rewrite to $\sqrt{a_n+6}

Now realize that $\sqrt{x+6}$ is continuous, so that, when setting $\lim a_n=L$, we know: $$L=\lim a_{n+1}=\lim \sqrt{a_n+6}=\sqrt{(\lim a_n)+6}=\sqrt{L+6}$$
Solving for $L$ yields $L\in\{-2,3\}$, and since a lower bound was $3$, we know $\lim a_n=3$.

Hope this helped!