 Summable increments imply a Cauchy sequence

If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.