The following is a proof of Weierstrass’ M-test:
Let $\{v_{n}\}$ be a sequence of continuous functions on the interval $[a, b]$, and assume that there is a convergent series $\sum_{n=1}^{\infty} M_{n}$ of positive numbers such that $|v_{n}(x)| \le M_{n}$ for all $n$. Then series $\sum_{n=1}^{\infty} v_{n}(x)$ converges uniformly on $[a, b]$.
Proof: Since $(C([a, b], \mathbb{R}), \rho)$ is complete, we only need to check that the partial sums $s_{n}(x) = \sum_{k=0}^{n} v_{k}(x)$ form a Cauchy sequence. Since the series $\sum_{n=0}^{\infty} M_{n}$ converges, we know that its partial sums $S_{n}=\sum_{k=0}^{n} M_{k}$ form a Cauchy sequence. Since for all $x \in [a, b]$ and all $m > n$, $$|s_{m}(x) − s_{n}(x)| = |\sum_{k=n+1}^{m}v_{k}(x)| \le \sum_{k=n+1}^{m} |v_{k}(x)| \le \sum_{k=n+1}^{m} M_{k} = |S_{m}-S_{n}|,$$
this implies that $\{s_{n}\}$ is a Cauchy sequence.
So, while I understand that the partial sums $S_{n}=\sum_{k=0}^{n} M_{k}$ form a Cauchy sequence as a result of convergence of $\sum_{n=0}^{\infty} M_{n}$, how do we know that $m > n$ are sufficiently large so that $|S_{m}-S_{n}|$ converges and is a Cauchy sequence?