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,

As @evinda rightly noticed, the limit must be $\frac12(1+\sqrt{13})$. For $a_1$ we can take an arbitrary number in $[-3,\infty)$. Note that $a_2$ will be nonnegative in any case and finally note that if $a_k\in [0,\frac12(1+\sqrt{13})]$, then $a_k\le a_{k+1}=\sqrt{a_k+3}\le \frac12(1+\sqrt{13})$ while if $a_k>\frac12(1+\sqrt{13})$, then $\frac12(1+\sqrt{13})<a_{k+1}=\sqrt{a_k+3}<a_k$; thus, the sequence is monotone starting from the second term and hence convergent.
