# Recurrence Relation - limit of a sequence

1. Sep 7, 2012

### dumbQuestion

Hello,

It is my understanding that if we have a sequence defined as follows:

an+1=(ψ)an + (λ)

Then if ψ≥1 or ψ≤-1, the sequence diverges. If -1<ψ<1, the sequence converges to:

λ/(1-ψ)

I was working problems in a book and one of the problems said that the following sequence converges:

xn=√(5xn-1+6), where x1=2

Can somebody explain to me why this converges? I mean, I guess it grows at such a slow rate that eventually it converges to a set limit. But what are some ways I can determine that limit? I mean are there any tools to evaluate the convergence of recurrance relations, like basic tools we have when dealing with regular sequences? The only "tool" I know is the one I listed above and my calculus book doesn't really look at these. Or when it comes to these, do I just have to plug in the first few numbers and look for a general trend? Thanks for the advice!

2. Sep 7, 2012

### haruspex

A useful trick is to suppose first of all that it does converge, to x say, then think what that would mean if you put xn-1 = x. You should then be able to calculate x (may be more than one solution). Some solutions may be attractors, i.e. if you start within a certain range - the basin of attraction - you will converge to that limit; others may not be, i.e. no matter how close you start to that solution you won't converge there. Knowing a potential limit allows you to study how the difference between a current value and the proposed limit changes with each step - does it shrink to zero?