The recurrence relation is as follows,
$$a_{n+1}=\frac{1+\sqrt{4a_{n}+9}}{2},a_0=3$$
On substituting values, I observed that the sequence appears to converge to 2.732, but as it is part of a larger problem, I would like to obtain a closed form of the limit.
I tried using generating functions, as well as other standard methods to try solving it but to no avail.
Any insight on the problem would be greatly appreciated.
Best Answer
If the limit exists, let's say $a_n \to L$, we must have $$ L=\frac 12 (1+\sqrt{4L+9}) \Leftrightarrow L = 1+ \sqrt{3}. $$
Of course, you must show that the limit actually exists. You can do this by elementary methods or you can use the fixed point theorem ($g(x)=\frac 12 (1+\sqrt{4x+9})$ is invariant and contractive in $[0,3]$).