Sequence problem I have a calculus final two days from now and we have a test example. There's a sequence question I can't seem to solve and hope someone here will be able to help.
With $a_1$ not given, what are the possible values of it so that the sequence $a_{n+1}=\sqrt{3+a_n}$ will converge. If it does, what is the limit?
I have no clue what so ever on what doing here. I mean, I can't prove the sequence is monotone. I assume that $a_1$ $\ge $ -3 and can also approach infinity.
Any help is appreciated,
Regards,

We want that the sequence $a_{n+1}$ converges, so $a_{n+1} \to l \in \mathbb{R} \Rightarrow a_n \to l$
Taking the limit $n \to +\infty$ at the relation $a_{n+1}=\sqrt{3+a_n}$ we get: $$l=\sqrt{3+l} \Rightarrow l^2=3+l \Rightarrow l=\frac{1}{2}(1-\sqrt{13}) \text{ OR } l=\frac{1}{2}(1+\sqrt{13})$$
As $\frac{1}{2}(1-\sqrt{13})$ is negative,we reject it,so the only possible limit of $a_n$ is $l=\frac{1}{2}(1+\sqrt{13})$.  
