Here's Theorem 3.22 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.
$\sum a_n$ converges if and only if for every $\varepsilon > 0$ there is an integer $N$ such that $$\left\vert \sum_{k=n}^m a_k \right\vert \leq \varepsilon $$ if $m \geq n \geq N$.
My proof:
Let $s_n = \sum_{k=1}^n a_k$ for each $n = 1, 2, 3, \ldots$. If $\sum a_n$ converges, then the sequence $\{s_n \}$ is Cauchy, so, given $\varepsilon > 0$, we can find an integer $N$ such that $$\left\vert s_m - s_n \right\vert < \varepsilon$$ if $m \geq N$, $n \geq N$. So if $m \geq n \geq N$, then we have $$ \left\vert \sum_{k=n}^m a_k \right\vert = \left\vert s_m - s_n \right\vert < \varepsilon. $$ Conversely, if for every $\varepsilon > 0$ there is an integer $N$ such that $$\left\vert \sum_{k=n}^m a_k \right\vert \leq \varepsilon$$ if $m \geq n \geq N$, then $\{s_n\}$ is a Cauchy sequence in $\mathbb{C}$ and therefore the series converges.
Note: In the converse part, we could have added a few steps to get the $<$ instead of the $\leq$, but that is not so important in this particular context. Am I right? a
