# 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.

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.

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

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