Iteration, linear function. convergence and divergence

1. Oct 20, 2011

janewaybos

1. The problem statement, all variables and given/known data
I need to understand and prove the following: That if a>1 the function diverges, except for a special case x_0= b/(1-a). Then if a=-1 diverges for some cases and converges if x_0 is b/2. Again, not to clear on this.

2. Relevant equations

lim n →∞ a^n(x_0+b/(a-1))-b/(a-1)

3. The attempt at a solution

For a>1, then we just need to look at x_0 and b/(a-1)? If x_0 > b/(a-1), then this is + and still diverges, then we rely on -b/(a-1)? I am kind of confused and any help would be appreciated.

Last edited by a moderator: Oct 20, 2011
2. Oct 20, 2011

Staff: Mentor

MOD Note: Is this your limit? What you wrote is hard to understand.
$$\lim_{n \to \infty} a^n \left(x_0 + \frac{b}{a - 1}\right) - \frac{b}{a - 1}$$

3. Oct 20, 2011

janewaybos

Yes.

4. Oct 21, 2011

HallsofIvy

??? For a equal to 1 you have $a^n= 1$ so the two fractions cancel, but not a not equal to 1.