The Cauchy Criterion for convergence states that if a series converges, it is both necessary and sufficient that
$$\forall \varepsilon \exists N:\forall n,m > N, \sum_{n}^{m}a_n < \varepsilon.$$
The contrapositive of this is:
$$\exists\varepsilon :\forall N, \exists n,m > N : \sum_{n}^{m}a_n \geq \varepsilon.$$
In the convergent case, its obvious that all $\varepsilon$ work. In the divergent case, is it reasonable to assume there could be multiple $\varepsilon$'s that work? A sketch of an idea would be that we can vary our interval [n,m] with $\varepsilon$ to find a suitable value for any $\varepsilon$, but this seems naive to me. Can anyone answer definitively or supply a better proof?
If $\sum_n^m a_n \geq \varepsilon > 0$, then $\sum_n^m a_n \geq \varepsilon'$ for all $\varepsilon' \in (0, \varepsilon)$. So all the $\varepsilon' \in (0, \varepsilon)$ work.