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.
Best Answer
$$ a_{n+1}^2-a_n^2=6+a_{n}-a_n^2=(3-a_{n})(2+a_n) $$
If $a_n>3, a_{n+1}>\sqrt{6+3}=3$. So by induction $a_n>3\;\forall\;n$ and $a_{n+1}^2<a_{n}^2\;\forall\;n$. And only possible limit is the positive solution of $x^2=6+x$.