# If $\sum_{n=1}^\infty\vert a_{n+k}-a_n\vert<\infty$, then $\{a_n\}_{n=1}^\infty$ has a convergent subsequence

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.

• The first inequality seems to be wrong. – Kavi Rama Murthy Dec 31 '20 at 9:27
• 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. – Steven Dec 31 '20 at 9:27
• @Jean-ClaudeArbaut Sorry, my idea was wrong, it is not a valid step – Stackman Dec 31 '20 at 9:33
• – 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.

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

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

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