1. ## 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?

2. ## Re: Convergence

You haven't specified what $b_{n+1}$ is. But you're on the right track - you can get a simple proof that works by showing that it's a Cauchy sequence.

3. ## Re: Convergence

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?

4. ## Re: Convergence

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:

$|a_{n+k} - a_{n}| = |(a_{n+k} - a_{n+k-1}) + (a_{n+k-1} - a_{n+k-2}) + ... + (a_{n+1} - a_{n})|$

$\le |a_{n+k} - a_{n+k-1}| + |a_{n+k-1} - a_{n+k-2}| + ... + |a_{n+1} - a_{n}|.$

5. ## Re: Convergence

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.
$$|a_{n+1}-a_n|<r^n$$ gives $|a_{n+1}-a_n|. Is that nice?
Can you show that $\left| {a_{n + 1} } \right| \leqslant \sum\limits_{k = 1}^n {r^k }~?$