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