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?