# cauchy sequences and their relation to sums of distances in metric spaces

The question is to prove or disprove (via counterexample) the following three statements:

a) A sequence $x_n$ is Cauchy iff $$\sum_{n=1}^{\infty} x_{n+1} - x_n$$ is convergent.

I think this one is true. By the Cauchy criterion for convergence of series, $\sum_{j=n}^{m} x_{j+1} - x_j = x_m - x_{n+1}$ converges iff $x_m - x_{n+1} \leq \epsilon$ where $\epsilon > 0$ is given. This would suggest $x_n$ is a Cauchy sequence.

b) If $x_n$ is Cauchy in a metric space $(M, d)$, then $$\sum_{n=1}^{\infty} d(x_{n+1}, x_n)$$ is convergent.

This one also seems to be true. If $x_n$ is Cauchy, then for $m, n \geq N$, $d(x_n, x_{n+1}) \leq \epsilon / (m - n)$ where $\epsilon > 0$ is given. Then $\sum_{j=n}^{m} d(x_{j+1}, x_j) \leq \sum_{j=n}^{m} \epsilon / (m - n) \leq \epsilon$, which would suggest convergence of the series as well. I'm not 100% sure about this proof, though.

c) If $x_n$ are points in a metric space $(M, d)$ such that $$\sum_{n=1}^{\infty} d(x_{n+1}, x_n)$$ is convergent, then $x_n$ is a Cauchy sequence in $(M, d)$.

• For b) consider a sequence where $d(x_{n+1}, x_n) = 1/n$. This is Cauchy but divergent. May 6, 2018 at 20:05
• @LukasKofler any chance you have an example? Would a recursive sequence suffice?
– user486635
May 6, 2018 at 20:11

I hope this is a complete and correct answer to all three parts. The series $\sum (x_{n+1}-x_n)$ converges if and only if the partial sums form a Cauchy sequence iff $\sum_k^{m} (x_{n+1}-x_n) \to 0$ as $k,m \to \infty$ iff $x_{m+1}-x_k$ $\to 0$ as $k,m \to \infty$ iff $\{x_n\}$ is cauchy. so a) is true. b) is false: if $\{e_n\}$ is an orthonormal set in an inner product space and $x_n=\frac {e_n} {n}$ then $d(x_n,x_{n+1}) \geq \sqrt \frac 2 {(n+1)^{2}}$ so $\sum d(x_n,x_{n+1})=\infty$. Of course, $x_n \to 0$ so $\{x_n\}$ is Cauchy. So b) is false. c) is true and teh proof involves just triangle inequality: $d(x_n,x_{n+m}) \leq \sum _n^{n+m-1} d(x_k, x_{k+1}) \to 0$

You proved one direction of (a) (the other isn't necessarily true), and (b)'s proof is incorrect, as the sequence $$x_n = \sum_{k=1}^{n} \left(\frac{-1}{k}\right)^k$$ the alternating harmonic numbers, disproves it.

For (c), as the sum is convergent, for any $\epsilon > 0$, there must exist an $N \in \mathbb{N}$ such that $$\sum_{k=1}^{\infty} d(x_k, x_{k+1}) - \sum_{k=1}^{N} d(x_k, x_{k+1}) = \sum_{k=N}^{\infty} d(x_k, k_{n+1}) < \epsilon$$ Clearly, for $m > n > N$, $$\sum_{k=n}^{m-1} d(x_k, x_{k+1}) < \sum_{k=N}^{\infty} d(x_k, x_{k+1})$$ And by the triangle inequality, $$d(x_n, x_m) \leq \sum_{k=n}^{m-1} d(x_k, x_{k+1})$$ Thus, $d(x_n, x_m) < \epsilon$, proving the sequence to be Cauchy.

• Your counterexample for (b) can't be using an increasing sequence in the reals, because that gives exactly the statement of (a). May 6, 2018 at 20:19
• One direction of (a) is true. I'm not sure why my counterexample doesn't hold.
– user211599
May 6, 2018 at 20:23
• The Harmonic numbers do not form a Cauchy sequence - else they are a convergent sequence. May 6, 2018 at 20:25
• Yea, my bad. It should be alternating.
– user211599
May 6, 2018 at 20:26
• What is the counterexample for the other direction of a?
– user486635
May 6, 2018 at 22:19