As we know, the Weierstrass function defined by $$ f(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x), 0<a<1,b \in 2\mathbb N+1,ab > 1+\frac{3\pi}{2} $$ is differentiable nowhere (see here for a proof of the same). As a consequence, $f$ is not of bounded variation on any finite interval, ($f$ is said to be of bounded variation on $[a,b]$ if for any partition $a=x_0<x_1<x_2<\cdots<x_{n-1}<x_n=b$, $\sum_{k=1}^n|f(x_k)-f(x_{k-1})|\leq M$ for a constant $M$ independent of the choice of partition) since functions of bounded variation are differentiable a.e. (as a corollary of this and the Lebesgue theorem). In particular, the proof that $f$ is not of bounded variation is not shown explicitly via the definition of functions of bounded variation.
Question: Can we construct a sequence of partitions on any interval $[a,b]$ for which the sequence of sums $\sum_{k=1}^n|f(x_k)-f(x_{k-1})|$ increases to positive infinite, showing that $f$ is not of bounded variation on $[a,b]$?