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.