Let $(a_n)$ be a sequence such that $\lim_{N\rightarrow \infty} \sum_{n=1}^N |a_n - a_{n+1}| < \infty$. Show that $(a_n)$ is Cauchy.

Proof: We know $(x_n)$ is Cauchy if $\forall\ \epsilon > 0, \exists\ N \in \mathbb{N}$ such that $m, n \geq N \implies |x_{m} - x_{n}| < \epsilon$.

Since we have that $\lim_{N\rightarrow \infty} \sum_{n=1}^N |a_n - a_{n+1}| < \infty$, and is the limit of the sum of absolute values (hence always positive and cannot diverge to $-\infty$), $\lim_{N\rightarrow \infty} \sum_{n=1}^N |a_n - a_{n+1}| = L$, for some $L \in \mathbb{R}$.

So we have that the limit of our sequence $|a_1 - a_{2}|, |a_1 - a_{2}| + |a_2 - a_{3}|\ + \dots$, converges.

Let $\epsilon > 0$. Take $x_n = \sum_{k=1}^n |a_k - a_{k+1}|$. Hence $\exists\ N \in \mathbb{N}$ such that $k \geq M \implies |x_k - x| < \frac{\epsilon}{2}$.

So $m , n \geq M \implies |x_m - x_n| = |(x_m - x) - (x_n - x)| \leq |x_m - x| + |x_n - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

So our sequence is Cauchy.

See comments, misread the question.

  • $\begingroup$ The proof is not complete. So far you deduced something about $x_n$ not $a_n$. $\endgroup$
    – xbh
    Oct 21, 2018 at 2:30
  • $\begingroup$ Completely misread the question and took the entire summation as the sequence, whoops. Never mind, not nearly this easy. $\endgroup$
    – SS'
    Oct 21, 2018 at 2:31
  • $\begingroup$ Don't worry. You are right on track. One more step would lead you to the goal. $\endgroup$
    – xbh
    Oct 21, 2018 at 2:32
  • $\begingroup$ $\sum_{n} |a_n - a_{n+1}|$ converges, hence $\sum_{n} (a_n - a_{n+1})$ converges, hence $a_n$ converges, hence $a_n$ is Cauchy. $\endgroup$ Oct 21, 2018 at 9:33

1 Answer 1


Picking up on your idea,

let $\epsilon>0$ and choose an integer $N$ so large that if

$m>N,\ \sum^\infty_{k=m}|a_k - a_{k+1}|<\epsilon. $

Then, if $n>m>N,$ we have

$ |a_n-a_m|=|\sum_{k=m}^{n-1} (a_{k+1}-a_k)|<\sum_{k=m}^{n-1} |a_{k+1} -a_k|=$

$\sum_{k=m}^{n-1} |a_k - a_{k+1}|<\sum_{k=m}^{\infty} |a_k - a_{k+1}|<\epsilon$


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