Originally Posted by

**simplysparklers** Hi guys

Can anyone put me right on this? I keep coming up with the opposite answer.

$\displaystyle a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3}$

$\displaystyle (n\geq{1})$

Show that $\displaystyle a_{n}$ is monotonic increasing.

Anyway I do it, I keep finding $\displaystyle a_{n}$ is monotonic decreasing. I've tried doing it as $\displaystyle a_{n+1}-a_{n}\geq{0}$, and I've gone very simple and just substituted different random values in for $\displaystyle a_{n}$ to see what kind of resuts it would yield, but it always ends up as $\displaystyle a_{n+1}\leq{a_{n}}$. Can anyone show me where I'm going wrong please??

Thanks,

Jo