Proof of convergence by proving a sequence is Cauchy

[KPutsch]

Homework Statement

Let 0 < r < 1. Let {A_n} be a sequence of real numbers such that |A_n+1 - A_n| < r^n for all naturals n. Prove {A_n} converges.

Homework Equations

A sequence of real numbers is called Cauchy, if for every positive real number epsilon, there is a positive integer N such that for all natural numbers m, n > N |x_m - x_n| < epsilon.

A sequence is convergent if and only if it is Cauchy.

Triangle Inequality: |x + y| <= |x| + |y|

The Attempt at a Solution

WLOG m>n, then |Am - An| = |Am - Am-1 + Am-1 - Am-2 ... -An|. Then we can use Triangle inequality to see that |Am - Am-1| < r^n, |Am-1 - Am-2| < r^n, ... . Then we can say that |Am - Am-1| + |Am-1 - Am-2| +...+ |An-1 -An|< n*r^n. Since r^n converges to zero, nr^n converges to zero, and nr^n < epsilon. Now to solve for n...

I have no idea how to solve nr^n < epsilon for n. Also, am I doing this right? I only have one poor example to go off of. If I can find n, will that mean I've shown that A_n is Cauchy?

Thank you for any help.

 

[JThompson]

Depends on the professor, but mine just used:
Let E>0 be given. Choose N so that for n>N, nr^n<E.
Didn't solve for N at all, since we already know that nr^n converges.

 

[ZioX]

You do not know n*r^n goes to zero. If you did you would be done.

|Am - Am-1| + |Am-1 - Am-2| +...+ |An-1 -An|< n*r^n

This is not true.

Also, since r is between 0 and 1 you know there is a natural number, say, a, such that 1/a <= r.

You know this because 1/n -> 0.

 

[KPutsch]

Ah, I proved in an earlier example that if 0 < r < 1, then nr^n converges to zero.

I realize now that the last pair of terms should be |An+1 -An|. The inequality should be true now: |Am - Am-1| + |Am-1 - Am-2| +...+ |An+1 -An|< n*r^n. So, am I done since I've shown that |An+1 - An| < ... < nr^n < epsilon? I really don't understand how this Cauchy sequence business works.

 

[ZioX]

No, it's not.

|A_m-A_n|\le |A_m-A_{m-1}|+\dots+|A_{n+1}-A_n|&lt;r^{m-1}+r^{m-2}+\dots+r^{n-1}&lt;(m-n)r^{n-1}

Where the last inequality follows from the fact that r^m<r^n for m>n and noting there are m-n terms summed.

Read some more Cauchy proofs. Pay particular attention to how m and n are handled. Remember that there exists some natural number eta s.t 1/eta <= r and for any pi > eta, 1/pi < 1/eta. Also, n is not a free variable in the sense that it is bounded by m.