The Cauchy Criterion of a series is as follow:
Theorem: Let $a_n$ be a real sequence. Then the infinite series $\sum_{k=1}^{+\infty}{a_k}$ converges if only if for every $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that $m > n > N$ implies
$$
\left| \sum_{k=n+1}^{m}{a_k} \right| < \varepsilon .
$$
From this result, how to demonstrate the following corollary?
Corollary: Let $a_n$ be a real sequence. Then the infinite series $\sum_{k=1}^{+\infty}{a_k}$ converges if only if for every $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that $n > N$ implies
$$
\left| \sum_{k=n+1}^{+\infty}{a_k} \right| < \varepsilon .
$$
Best Answer
By Cauchy's Theorem, for every $\varepsilon > 0$ there is $N \in \mathbb{N}$ such that $n > N$ imply $$ \left| \sum_{k = n + 1}^{n+p}{a_k} \right| < \varepsilon, $$ for all integer positive $p$. Let $S_{p}^{(n)}$ represent the sequence of partial sums of $\sum_{k = n + 1}^{n+p}{a_k}$, when $ n > N $ is fixed. Note that by the Principle of Mathematical Induction that $S_{p}^{(n)}$ is true for all $p \in \mathbb{N}$. Thus, $$ \left| \sum_{k = n + 1}^{+\infty}{a_k} \right| < \varepsilon. $$
Conversely, let $m, n $ be positive integers such that $m > n > N$. Thus,
$$ \left| \sum_{k = n + 1}^{m}{a_k} \right| \leq \left| \sum_{k = n + 1}^{+\infty}{a_k} \right| < \varepsilon. $$
Hence, by the Cauchy Criterion, $\sum{a_k}$ converges.