Problem: Let $k\in\mathbb N$ be fixed and suppose that the series $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert$ converges. Prove that $\{a_n\}_{n=1}^\infty$ has a convergent subsequence.

My Thoughts: Using the triangle inequality we see that for any fixed $n\in\mathbb N$ with $n>k+1$ we have \begin{align*} \sum_{j=k+1}^{n}\vert a_{j+k}\vert-\sum_{j=1}^k\vert a_j\vert &\leq \sum_{j=1}^n\vert a_{j+k}-a_j\vert\\ &\leq \sum_{j=1}^\infty\vert a_{j+k}-a_j\vert\\&\leq M, \end{align*} for some $M>0.$ Therefore we have the uniform bound $$\sum_{j=k+1}^{n}\vert a_{j+k}\vert\leq M+\sum_{j=1}^k\vert a_j\vert,$$ and hence the series on the left-hand side converges. This implies that $$\lim_{n\to\infty}a_{n+k}=0,$$ and we have our convergent subsequence.

Do you agree with the proof presented above?
Thank you for your time and feedback.

  • 2
    $\begingroup$ The first inequality seems to be wrong. $\endgroup$ – Kavi Rama Murthy Dec 31 '20 at 9:27
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    $\begingroup$ Your conclusion is that there is a subsequence converging to zero. But evidently the sequence $a_n := 1$ satisfies your hypotheses, so your argument can't be right. $\endgroup$ – Steven Dec 31 '20 at 9:27
  • $\begingroup$ @Jean-ClaudeArbaut Sorry, my idea was wrong, it is not a valid step $\endgroup$ – Stackman Dec 31 '20 at 9:33
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    $\begingroup$ Related: math.stackexchange.com/q/277152/42969. $\endgroup$ – Martin R Dec 31 '20 at 9:40

Let $M= \sum\limits_{n=1}^{\infty} |a_{n+k}-a_n|$. Check that $ \sum\limits_{n=1}^{\infty} |a_{k(n+1)}-a_{kn}| \leq M$. The sequence $b_n=a_{k n}$ satisfies $\sum |b_{n+1}-b_n| <\infty$. This implies that $(b_n)$ is Cauchy.


Consider the sequence $(b_n) = (a_{kn})$. From your hypothesis one can deduce that this sequence is a Cauchy sequence, so that it must be convergent.

Let $\epsilon > 0$. Since $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert$ converges, there is a $N$ such that $\sum_{n=N}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon.$ Since $k\geq 1,$ we also have $\sum_{n=kN}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon.$

Now let $m_2>m_1>N.$ Then

$$ \vert b_{m_2} - b_{m_1} \vert = \vert a_{km_2} - a_{km_1} \vert = \vert (a_{km_2} - a_{k(m_2-1)}) + \ldots + (a_{k(m_1+1)} - a_{km_1})\vert \leq {\vert a_{km_2} - a_{k(m_2-1)} \vert + \ldots + \vert a_{k(m_1+1)} - a_{km_1}\vert} \leq \sum_{n=kN}^{\infty} \vert a_{n+k}-a_n\vert < \epsilon. $$

$(b_n)$ is therefore a Cauchy sequence and hence convergent.

  • $\begingroup$ My very first thought was to show that sequence is Cauchy, but I didn't see how to do it. $\endgroup$ – Adam Rubinson Dec 31 '20 at 9:41
  • $\begingroup$ @AdamRubinson I wrote out the details, hope it's clear enough. $\endgroup$ – Steven Dec 31 '20 at 9:54

Since absolute convergence implies convergence,

$\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert<\infty \implies \sum_{n=1}^\infty\left( a_{n+k}-a_n\right)<\infty.$

Next, if it weren't true that $ \sum\limits_{n=1}^{\infty} \left(a_{k(n+1) + i}-a_{kn+i} \right) $ converges for all $i \in \{0,...,k-1 \}$, then the original sum $\sum_{n=1}^\infty\left( a_{n+k}-a_n\right)$ would not converge, a contradiction.

So we choose one $i \in \{0,...,k-1 \}$, and consider the subsequence $\{b_n\} = \{ a_{kn+i} \} $ of $\{a_n\}.$

We already have that $ \sum\limits_{n=1}^{\infty} \left(b_{n+1}-b_n\right) = L$, so I'll show that {$b_n$} is a convergent sequence.

By supposition, $ \sum\limits_{n=1}^{\infty} \left(b_{n+1}-b_n\right) = L \in \mathbb{R}.$

By cancellation of terms, this is the same as:

$lim_{n \to \infty} \left(b_{n+1}-b_1\right) = L$

$\implies lim_{n \to \infty} \left(b_{n}\right) = L + b_1.$


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