So I've been been working on the question below, and I have some questions in regards to the validity of my answer.

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

I make the claim that the distance between the terms of $(a_n)$ must approach zero. As such for every $\epsilon > 0$, there must be be an integer $N$ such that

$$|a_m - a_n| < \epsilon$$

for all $m,n \geq N$. That is, the sequence is Cauchy. To show this, assume that the distances between the terms of $(a_n)$ do not approach zero. Let

$$a=\min \left \{ |a_1 - a_2|,|a_2 - a_3|,...,|a_N - a_{N+1}|,... \right \}.$$

Then we have $a \neq 0$. Observe that

$$\lim_{N \to \infty} Na \leq \lim_{N \to \infty} \sum_{n=1}^{N} |a_{N}-a_{N+1}| < \infty,$$

which, is a contraction, as

$$\lim_{N \to \infty} Na= \infty$$

for any $a \neq 0$. Thus, we must have $a=0$, and the distance between the terms of $(a_n)$ must approach zero and as such the sequence is Cauchy.

I am unsure about setting $a= \min \{...\}$. Any input or comments about my answer would be appreciated.

  • $\begingroup$ What if you change $a_2$ to be the same as $a_1$ keeping all other terms the same. The convergence properties will not change, but $a$ will be 0. $\endgroup$ – Arin Chaudhuri Feb 21 '13 at 17:20

Your $a$ could be $0$, so this does not work, even if the idea is great.

Take $\epsilon>0$.

Since the series converges, there exists $N$ such that $$ \sum_{n=N}^{+\infty} |a_{n+1}-a_n|\leq \epsilon. $$

Now for all for $N\leq k<l$, we have $$ |a_k-a_l|=|\sum_{n=k}^{l}a_{n+1}-a_n|\leq \sum_{n=k}^{l}|a_{n+1}-a_n|\leq \sum_{n=N}^{+\infty}|a_{n+1}-a_n|\leq \epsilon. $$

So the sequence $a_n$ is indeed Cauchy.


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