A sequence $\{a_n\}$ has bounded variation if $\sum_{k=1}^\infty |a_{k+1}-a_k|$ converges. I am trying to prove that, if $\{a_n\}$ has bounded variation then $\sum_{k=1}^\infty a_n$ converges. This is for an exercise from Kosmala's real analysis textbook, (7.4 problem 20 (c)).
It's in the section which covers alternating series, although I don't see a way to relate this to an alternating series. My best guess is that it is meant to be somehow related to absolute convergence of the of the bounded variation condition.
I've tried looking at the Cauchy condition,
$$\left|\sum_{k=m}^n a_k\right| < \varepsilon$$
and thought of how one could try to relate this to the bounded variation condition. Maybe
$$ \left|\sum_{k=m}^n (a_k-a_{k+1}+a_{k+1}) \right| = \left|\sum_{k=m}^n (a_k-a_{k+1})+\sum_{k=m}^n a_{k+1}\right|$$
$$\le \sum_m^n |a_{k+1}-a_k|+\left|\sum_m^n a_{k+1}\right|$$
The first term after the inequality must go to zero eventually, but this looks like a complete dead-end to me.
Note: my question is not about the convergence of $\{a_k\}$ and therefore is not related to the following question: Prove a sequence with bounded variation converges. There seems to have been some confusion over this.
You cannot prove it, since it is false. For instance, the harmonic series ($\sum_{n=1}^\infty\frac1n$) diverges, in spite of the fact that the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is a sequence of bounded variation.