Let $\Omega \subseteq \mathbb C$ be a region and $f : \Omega \longrightarrow \mathbb C$ be a continuous function. Let $\gamma : [0,1] \longrightarrow \Omega$ be a continuous path of bounded variation. For $n \in \mathbb N,$ define $$b_n = \sum\limits_{k=0}^{2^n - 1} f \left (\gamma \left (\frac {k} {2^n} \right ) \right ) \left (\gamma \left (\frac {k+1} {2^n} \right ) - \gamma \left (\frac {k} {2^n} \right ) \right ).$$ Show that $\{b_n\}_{n \geq 1}$ is a Cauchy sequence.
$\textbf {My Attempt} :$ What I found is that for any $n \in \mathbb N$ we have $$b_{n+1} - b_n = \sum\limits_{\substack {k=0} \\ k\ \text {odd}}^{2^{n+1} - 1} \left [f \left (\gamma \left (\frac {k} {2^{n+1}} \right ) \right ) - f \left (\gamma \left (\frac {k-1} {2^{n+1}} \right ) \right ) \right ] \left (\gamma \left (\frac {k+1} {2^{n+1}} \right ) - \gamma \left (\frac {k} {2^{n+1}} \right ) \right ).$$ From here I am trying to estimate $|b_{n+p} - b_n|$ for any natural number $p.$ Since $\gamma$ is a function of bounded variation it follows that the total variation $V_{[0,1]} (\gamma) = M \lt \infty.$ Since $f \circ \gamma$ is continuous on $[0,1],$ it is uniformly continuous there. Then for a given $\varepsilon \gt 0$ there exists $\delta \gt 0$ such that for any pair of points $x,y \in [0,1]$ with $|x - y| \lt \delta$ we have $|f(\gamma (x)) - f(\gamma(y))| \lt \frac {\varepsilon} {M}.$ Choose $n_0 \in \mathbb N$ with $\frac {1} {2^{n_0 + 1}} \lt \delta.$ Then for all $n \geq n_0$ we have $|b_{n+1} - b_n| \lt \varepsilon.$ But that is not the thing I want. Rather I want $|b_{n+p} - b_n| \lt \varepsilon$ after a certain stage for any $p \in \mathbb N.$ How do I modify my proof to reach at the desired conclusion?
Any help in this regard would be warmly appreciated. Thanks for your time.
You may write
$$ b_{n+p} - b_n = \sum_{k=0}^{2^{n+p}-1} \biggl[ (f\circ\gamma)\biggl(\frac{k}{2^{n+p}} \biggr) - (f\circ\gamma)\biggl(\frac{2^p \lfloor k/2^p\rfloor}{2^{n+p}} \biggr) \biggr] \cdot\biggl[ \gamma\biggl(\frac{k+1}{2^{n+p}} \biggr) - \gamma\biggl(\frac{k}{2^{n+p}} \biggr) \biggr] $$
and then proceed just like in the original attempt.