Retrospective sequence

Question

Prove that the sequence which defined by $a_1=1$ and $a_{n+1}=\sqrt{12+4a_n}$ for every $n\geq1$ is convergent and find its limit.

I think the way to prove this is to show that $a_n$ is monotone and bounded. We can show that is monotone and increasing by induction. But i'm stuck at the bounded part.

Answer 1

To prove that the sequence converges, you can prove that it's bounded above and increasing:

1. Using induction, you can prove $a_n$ is bounded above by $6$. $$ a_n<6\Leftrightarrow 4a_n<24\Leftrightarrow 12+4a_n<36\Leftrightarrow \sqrt{12+4a_n}<\sqrt{36}\Leftrightarrow a_{n+1}<6 $$

2. Again using induction, you can prove that $a_n$ is increasing.

$$ a_{n-1}<a_n\Leftrightarrow 12+4a_{n-1}<12+4a_n\Leftrightarrow \sqrt{12+4a_{n-1}}<\sqrt{12+4a_n}\Leftrightarrow a_n<a_{n+1} $$

The limit can be found simply by replacing $a_i$ with $L$: $$ L=\sqrt{12+4L}\Leftrightarrow L\ge 0~~\land~~L^2-4L-12=0\Leftrightarrow L\ge 0~~\land~~(L+2)(L-6)=0 $$ $$ \Leftrightarrow L\ge 0~~\land~~L \in \{-2, 6\}\Leftrightarrow L=6 $$

Answer 2

You calculate the limit first, using $L=\sqrt{12+4L}$, yielding $L=6$. Now assume $a_n=6-\epsilon$, with $\epsilon>0$. Plug it into your reccurence relationship and show that $a_{n+1}<6$

Answer 3

You need to find a number $M$ with the property that $x \le M$ implies $\sqrt{12 + 4x} \le M$. As long as $x \ge -3$ it suffices to find the least positive solution $M$ to $\sqrt{12 + 4M} \le M$, which happens to be $6$.

Clearly $1 \le 6$, and observe $x_n \le 6$ implies $\sqrt{12 + 4x_n} \le \sqrt{12 + 24} = \sqrt{36} = 6$.

A bit more formally, you can define $E = \{n \in \mathbb N : -3 \le x_n \le 6\}$ and show that $1 \in E$ and $n \in E$ implies $n+1 \in E$ too. This avoids any issues surrounding the root.