Is $x_n = \sum_{i \in \mathbb{N}, \frac{m+ik-1}{n-1} \in [0,1]} |f(\frac{m+ik-1}{n-1})- f(\frac{m+(i-1)k-1}{n-1})|$ convergent when $f\in C\cap BV$? Assume $f$ is continuous and of bounded variation on the interval $[0,1]$, $f \in C[0,1] \cap BV[0,1]$. Then the sequence
$$
x_n = \sum_{i \in \mathbb{N}, \frac{m+ik-1}{n-1} \in [0,1]} \left | f(\frac{m+ik-1}{n-1})- f(\frac{m+(i-1)k-1}{n-1})\right|
$$
with $m,k \in \mathbb{N}, m \leq k$ is bounded by the variation of $f$. 
Can we show that the sequence is in fact convergent?
 A: Theorem: If $f$ is continuous on $[0,1],$ then
$$\lim_{|P|\to 0} \sum_{P}|\Delta f| = V_0^1(f).$$
Here $P$ is a partition of $[0,1],$ $|P|$ denotes the mesh size of $P,$ and $V_0^1(f)$ is the total variation of $f$ on $[0,1]$ (which might be $\infty$). I hope the notation $\sum_{P}|\Delta f|$ explains itself. 
The theorem implies the following solution to your question: With your hypotheses, $\lim_{n\to \infty}x_n = V_0^1(f).$ To see this, fix $m,k$ and define the partitions
$$P_n = \left (\{\frac{m+ik-1}{n-1} : i\in \mathbb N\}\cap [0,1]\right) \cup \{0,1\}.$$
Check that $|P_n|$ is on the order of $1/n$ as $n\to \infty.$ Thus by the theorem,
$$\tag 1 \lim_{n\to \infty} \sum_{P_n}|\Delta f| = V_0^1(f).$$
Now the sums in $(1)$ could be slightly larger than the sums in your problem, due to the fact that $0,1$ may not be among the sampling points that you specify. That's no problem. The most we could be adding is
$$\tag 2|f((a_n)-f(0)| + |f(1)-f(b_n)|,$$
where $a_n$ is the first of your points, and $b_n$ is the last of your points. Because $|a_n-0|,|1-b_n|=O(1/n),$ the continuity of $f$ shows the contribution in $(2)$ disappears in the limit. This proves $x_n \to V_0^1$ as desired.
Have you previously encountered the theorem I used?

Added later: Proof of the Thm: I'll prove it for the case $V_0^1(f)<\infty.$ Let $\epsilon>0.$ Then there is a partition $P=\{0=x_0<x_1<\cdots <x_n=1\}$ such that
$$\sum_{P}|\Delta f| > V_0^1(f)-\epsilon.$$
Let $I_k=(x_{k-1},x_k).$ If $\delta>0$ is small enough, then i) any partition $Q$ with $|Q|<\delta$ will contain at least two points in every $I_k,$ and ii) $|x-y| < \delta$ implies $|f(x)-f(y)| <\epsilon/(2n).$ Property ii) follows from the uniform continuity of $f$ on $[0,1].$
Fix such a $\delta,$ and assume $|Q|<\delta. $ I'll show $\sum_{Q}|\Delta f|>V_0^1(f)-2\epsilon.$ That will prove the theorem.
For $k=1,\dots,n,$ define $Q_k= Q\cap I_k.$ Since each $Q_k$ has at least two points, we can define $a_k= \min Q_k,$ $b_k = \max Q_k.$ Let $R$ be the partition $R=P\cup (Q_1 \cup \cdots \cup Q_n).$ Because $R$ is a refinement of $P,$ we have
$$\sum_{R}|\Delta f| \ge \sum_{P}|\Delta f| > V_0^1(f)-\epsilon.$$
Write 
$$\sum_{R}|\Delta f| = \sum_{k=1}^{n}\sum_{Q_k}|\Delta f| + \sum_{k=1}^{n}(|f(a_k)-f(x_{k-1})| + |f(x_k)-b_k|).$$
Each summand in the second sum on the right is $<2\epsilon/(2n)$ by ii) above, hence that sum $<\epsilon.$ It follows that the first sum on the right is at least $\sum_{R}|\Delta f|-\epsilon.$ Since $\sum_{Q}|\Delta f|\ge  \sum_{k=1}^{n}\sum_{Q_k}|\Delta f|,$ we're done.
