Help understanding proof of "$\sum |a_n|$ converges $\Rightarrow$ $\sum a_n$ converges" I'm self-studying Real Analysis. I can't understand a proof of this commonly known theorem:

Theorem. Every absolutely convergent series is convergent.
Proof. If $\sum |a_n|$ converges, given an arbitrary $\varepsilon>0$, there exists $n_0 \in \mathbb{N}$ such that $n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$, whichever $p \in \mathbb{N}$. So $|a_{n+1} + ... + a_{n+p}| \leq |a_{n+1}| + ... + |a_{n+p}|<\varepsilon$. Therefore, $\sum a_n$ converges by virtue of the Cauchy Criterion for series.

The passage that I don't understand is
$$n>n_0 \Rightarrow |a_{n+1}| + ... + |a_{n+p}|<\varepsilon.$$
Why is that the case? It doesn't seem obvious... and it doesn't seem to follow from any previous discussion. I don't know what I'm missing here.
 A: The series $\sum_{n=1}^{\infty}a_{n}$ is defined to converge if, for all $\epsilon>0$, there exists an $N\in\mathbb{N}$ such that $\left|\sum_{n=p}^{q}a_{n}\right|<\epsilon$ for all $N\leq p<q$. If, in particular, $\sum_{n=1}^{\infty}|a_{n}|$ converges, then given $\epsilon>0$, there exists (by definition) $N\in\mathbb{N}$ such that
\begin{equation*} \left|\sum_{n=p}^{q}|a_{n}|\right|=|a_{p}|+\ldots+|a_{q}|<\epsilon \end{equation*}
if $N\leq p<q$ but, of course,
\begin{equation*} \left|\sum_{n=p}^{q}a_{n}\right|\leq |a_{p}|+\ldots+|a_{q}|<\epsilon \end{equation*}
so $\sum_{n=1}^{\infty}a_{n}$ is convergent (by definition) as well.
A: Here provides another method:
Note that
$$a_n=\frac{|a_n|+a_n}{2}-\frac{|a_n|-a_n}{2},n\in \mathbb N.$$
Obviously, we have
$$0\leq\frac{|a_n|+a_n}{2}\leq|a_n|\ \ \text{and}\ \ 0\leq\frac{|a_n|-a_n}{2}\leq|a_n|,$$
by comparison test of positive series，both of the series
$$\sum_{n=1}^{\infty}\frac{|a_n|+a_n}{2}\ \ \text{and}\ \ \sum_{n=1}^{\infty}\frac{|a_n|-a_n}{2}$$
converge.
So the series $$\sum_{n=1}^{\infty}a_n$$
converges.
