Prove sequence in a complete metric space converges if the series of distances converges.

Question

Assume that X is complete, and let $(p_n)$ be a sequence in X. Assume that $\sum\limits_{n=1}^\infty d(p_n, p_{n+1})$ converges. Prove that $(p_n)$ converges.

(X,d) is a metric space.

Don't quite know how to start. Any help/hints appreciated!

Answer 1

1. How to start: look at the definition of "complete metric space". Every Cauchy sequence converges. OK, we have to show that $(p_n)$ is a Cauchy sequence. Using, somehow, the fact that the series of distances converges.

2. Next, consider the definition of a Cauchy sequence. Given a positive $\varepsilon$ we have to show that, if $n$ is big enough, then $d(p_n,p_m)\lt\varepsilon$ for all $m\gt n$. Right? Is that how the Cauchy criterion is stated in your book?

3. Hmm. How do we get from $d(p_n,p_{n+1})$ to $d(p_n,p_m)$?? The triangle inequality!! If $m=n+k$ then$$d(p_n,p_m)=d(p_n,p_{n+k})\le d(p_n,p_{n+1})+d(p_{n+1},p_{n+2})+\cdots+d(p_{n+k-1},p_{n+k})\le d(p_n,p_{n+1})+d(p_{n+1},p_{n+2})+d(p_{n+2},p_{n+3})+\cdots$$

4. Hmm. The remainder of a convergent series. Goes to zero, doesn't it?

Answer 2

HINT: Show that if $\sum_{n\ge 1}d(p_n,p_{n+1})$ converges, then $\langle p_n:n\in\Bbb Z^+\rangle$ is a Cauchy sequence. If that’s not quite enough, there’s a further hint in the spoiler-protected block below.

Further HINT: Show that the hypothesis implies that $\sum_{n\ge m}d(p_n,p_{n+1})\to 0$ as $m\to\infty$.