Thread: Convergence

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?

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.

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?

Originally Posted by lovesmath

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...
Does this work?

That won't work to prove convergence, since you're assuming convergence.

Here's a hint for setting it up to show that it's a Cauchy sequence:

Originally Posted by lovesmath

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.

@lovesmath why don't you learn to make your posts readable?
LaTeX is not that difficult.
[tex] |a_{n+1}-a_n|<r^n [/tex] gives . Is that nice?

Can you show that