Iteration, linear function. convergence and divergence

[janewaybos]

Homework Statement

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.

Homework Equations

lim n →∞ a^n(x_0+b/(a-1))-b/(a-1)

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.

 

[Mark44]

Homework Statement

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

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.

Homework Equations

lim n →∞ a^n(x_0+b/(a-1))-b/(a-1)

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.

 

[janewaybos]

Yes.

 

[HallsofIvy]

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