# Consider the sequence defined: $a_1=0, a_{n+1}=3+\sqrt{11+a_n}$, show that is bounded above and increasing using induction.

Consider the sequence defined: $a_1=0, a_{n+1}=3+\sqrt{11+a_n}$ a) Show, using induction, that this sequence is bounded above by 14; b) prove that the sequence is increasing; c) Why must it converge?; d)Find the limit.

So for part a), I have:

Let $P(n)$ be the statement that $a_n \leq 14$. Consider $P(1)$, where $n=1$. Then $0 \leq 14$, which is indeed true. Now assume that $P(k)$ is true for some $k\in \mathbb{N}$, i.e. $a_k \leq 14$ Then for $P(k+1)$: $a_{k+1}=3+\sqrt{11+a_k}=3+\sqrt{11+14}=8 \leq 14$. Thus the statement holds for $P(k+1)$ is true. Therefore the statement holds for all $n \in \mathbb{N}$.

For part b), I also use induction.

Let $S(n)$ be the statement: $a_{n+1} \geq a_n, \forall n \in \mathbb{N}$. Then for $n=1$, we have that $S(1)$ is $a_2=3+\sqrt{11} \geq a_1=0$, which is indeed true. Assume $S(k)$ is true for some $k \in \mathbb{N}$, i.e. $a_{k+1} \geq a_k$. Then for $S(k+1)$, $a_{k+2}=3+\sqrt{11+a_{k+1}} \geq 3+\sqrt{11+a_k}=a_{k+1}$. Thus $S(k+1)$ is true. Thus $S(n)$ is true for all $n \in \mathbb{N}$

c) The sequence must converge since this sequence is bounded and monotonically increasing. d) For this part do I say that $L=3+\sqrt{11+L}$, and then solve for $L$?

I just want to see if I am doing this correctly. I think that part b is incorrect because of the induction step. Any help and feedback is appreciated. Thanks in advance.

• It's all fine. Just one minor thing (probably a typo): you should have a $\leq$ in part a) where you wrote $3+\sqrt{11+a_k}=3+\sqrt{11+14}$. And yes finish off by solving for L. – user39572 Oct 2 '12 at 1:52
• Seems basically fine. In proof that the sequence is bounded, you want $a_{k+1}=3+\sqrt{11+a_k}\le 3+\sqrt{11+14}=8\le 14$. And you should say that the statement $P(k+1)$ is true. Your version not clear, "for" confusing. In second induction, there should be no $\forall n$ at the start. Rest of (b) is fine. – André Nicolas Oct 2 '12 at 1:55

A minor point: You find using your method that the limit $L$ is a solution to the quadratic equation $x^2-7x-2$. One of the solutions to this equation is positive, and the other is negative, and you can deduce that the limit is the positive solution.