I'm trying to understand step 3 of the proof of Merten's theorem.
How is it known that $N*_{sup}|B_i-B|$ term converges slow enough that there is an N where equation 3 is true: $|a_n|\le\frac{\varepsilon}{3N(\sup_{ i\in\{0,\dots,N-1\} } |B_i-B|+1)}$
Both $a_n$ and $|B_i-B|$ are converging, but the relative rate I don't understand.
Note that $N$ is defined for in the step before, in view of getting (2). In the third step, $N$ is now a fixed integer: so, defining $\varepsilon'>0$ as $$ \varepsilon' \stackrel{\rm def}{=} \frac{\varepsilon}{3N(\sum_{i < N}\lvert B_i - B\rvert +1)} $$ we have that $\varepsilon'$ is "just a number": so since $(a_n)_n$ converges to 0, there exists some $M\geq 0$ such that $$\lvert a_n \rvert \leq \varepsilon', \qquad \forall n \geq M$$ There is no issue of "relative rate of convergence": $N$ is fixed based on the convergence of $(B_n)_n$, then once $N$ is fixed $M$ is defined based on the convergence of $(a_n)_n$.