Proof that if the sum of the differences of consecutive terms of a sequence between any two terms is bounded above by one, then the sequence converges

Question

Someone asked me to help them with the following real analysis problem and it's been bugging me that I haven't been able to figure it out or think of something new to try for about a day now:

Suppose it holds for a sequence $\{a_j\}$ of complex numbers that for every pair of integers $N>M>0$, $$\lvert a_M-a_{M+1}\rvert+\lvert a_{M+1}-a_{M+2}\rvert+\ldots+\lvert a_{N-1}-a_{N}\rvert\le1.$$ Prove that $\{a_j\}$ converges.

It was clear to both of us that we have to show that the sequence is Cauchy. I tried contradiction a few times, though I (apart from negating the definition incorrectly a few times) couldn't begin to place where the contradiction would arise. I get that we must have $\lvert a_M-a_N\rvert\le1$, but can't see how that would imply that it's arbitrarily small. After thinking about it on my own, I realised that the smallest difference between consecutive terms must be no greater than $\frac{1}{N-M}$, but the largest one could theoretically be arbitrarily close to $1$, so I don't really see how that's helpful.

And that's about where I've been stuck for about a day. If anyone can give me hints about where to go, I'd appreciate it.

Also, yes, despite the exercise concerning a complex sequence, it is for a real analysis course.

Answer 1

Firstly, the sequence $(a_n)$ is bounded since \begin{align*} |a_n -a_1| &= |(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \cdots + (a_2 - a_1)|\\[4pt] &\le |a_n - a_{n-1}| + |a_{n-1} - a_{n-2}|+ \cdots + |a_2 - a_1|\\[4pt] &\le 1\\[4pt] \end{align*} It follows that the sequence $(a_n)$ has at least one limit point.

Suppose the sequence $(a_n)$ has two distinct limit points $u,v$, say. Let $\theta = |u-v|$.

Construct a subsequence $b_1,b_2,b_3,...$ of the sequence $(a_n)$ such that

• $|b_n - u| < {\large{\frac{\theta}{3}}}$ for all odd $n$.$\\[12pt]$
• $|b_n - v| < {\large{\frac{\theta}{3}}}$ for all even $n$.

It follows that $|b_{n+1}-b_n| > {\large{\frac{\theta}{3}}}$, for all n.

For each $n$, there are positive integers $j,k$, with $j < k$ such that $b_n = a_j$ and $b_{n+1} = a_k$. Then $$\frac{\theta}{3} < |b_{n+1}-b_n| = |(a_k-a_{k-1}) + \cdots + (a_{j+1}-a_j)| \le |a_k-a_{k-1}| + \cdots + |a_{j+1}-a_j| $$ But then, summing the LHS for all $n$ yields $\infty$, whereas summing the RHS for all $n$ yields at most $1$, contradiction.

It follows that the sequence $(a_n)$ can't have two distinct limit points, hence, since the sequence $(a_n)$ is bounded, it must converge.

Answer 2

Your $$\lvert a_M-a_{M+1}\rvert+\lvert a_{M+1}-a_{M+2}\rvert+\ldots+\lvert a_{N-1}-a_{N}\rvert\le1$$ means that the partial sums of the positive series $$\sum^\infty_{n=1}|a_n-a_{n+1}|$$ are bounded, i.e. the series converges. But that means that $$\sum^\infty_{n=M}|a_n-a_{n+1}|\to0$$ as $M\to\infty,$ and $$|a_N-a_M|\le\sum^N_{n=M}|a_n-a_{n+1}|\le\sum^\infty_{n=M}|a_n-a_{n+1}|$$ for $N>M,$ so it is a Cauchy sequence.
Alternatively, you could say that $$\sum^\infty_{n=1}(a_n-a_{n+1})$$ is convergent, because it converges absolutely, and its sum is $$a_1-\lim_{n\to\infty}a_n.$$

Answer 3

Write $A_N$ for $\sum_{n=1}^N|a_{n}-a_{n+1}|$. Can you show that $A_N$ converges, and hence is a Cauchy sequence?