# Sum of the distance between consecutive terms implies the sequence is Cauchy.

If a sequence $(x_n)_{n=1}^{\infty}$ in $\mathbb{R}^n$ satisfies $\sum_{n\geq 1} ||x_n-x_{n+1}||<\infty$, show that it is Cauchy.

This isn't a complete answer, but here's my train of thought. Since $\sum_{n\geq 1} ||x_n-x_{n+1}||$ is finite, $\lim_{n\to\infty}||x_n-x_{n+1}|| = 0$. This implies that $(x_n)$ is a monotone decreasing sequence. Next, we can find an upper bound for $x_n$. After showing that $x_n$ is bounded, we know it is a bounded monotone sequence so it converges and is therefore a Cauchy sequence.

I know I would still need to prove that there is an upper bound, but does this generally make sense of have I missed something? Thanks!

• For a vector-valued sequence, speaking of "monotonicity" makes little sense. Instead, try to use the triangle inequality applied to $\|x_n-x_{n+h}\|$. – Giuseppe Negro Oct 9 '14 at 23:34

Use that $\|x_{n+k}-x_n\|\le \|x_{n+k}-x_{n+k-1}\|+\cdots +\|x_{n+1}-x_n\|.$
Now, use that $$\sum_{n\geq 1} ||x_n-x_{n+1}||<\infty$$ to have control on the right part of the inequality, from where you have control on the left part.
Fix $\varepsilon > 0$. Then, there exists $N$ such that $$\sum_{k \geq N} \| x_k - x_{k+1} \| < \varepsilon.$$ Suppose that $m > n \geq N$. Then, why is it true that $\|x_m - x_n \| < \varepsilon$?