- #1

KPutsch

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## 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.