You haven't specified what is. But you're on the right track - you can get a simple proof that works by showing that it's a Cauchy sequence.
Suppose 0<r<1 and |a(sub n+1)-a(sub n)|<r^n for all n elements of the natural numbers. Prove that {a(sub n)}n=1 to infinity converges. Here is my work:
Since 0<r<1, then r^n --> 0. Since |a(sub n+1)-a(sub n)|<r^n for all n elements of the natural numbers, then this implies that |b(sub n+1)-a(sub n)| --> 0. Therefore, a(sub n) is Cauchy, and therefore must converge.
I feel like all I have done is repeat what was given in the problem. Is there something else I need to add or clarify?
Since r^n --> 0, then |a(sub n+1)-a(sub n)|<0 for all n elements of the natural numbers. Let E>0. Since a(sub n) --> A, there exists n in the natural numbers such that |a(sub n)-A|<E/2 whenever n>=N. If m,n>=N, then |a(sub n)-a(sub m)|<=|a(sub n)-A|+|a(sub m)-A|<E/2+E/2=E.
Does this work?