Absolute convergence of $\sum_{n=0}^{\infty}a_n\implies\sum_{n=0}^{\infty}a_n^2$ is absolute convergent too I am pretty sure that

Absolute convergence of $\sum_{n=0}^{\infty}a_n\implies\sum_{n=0}^{\infty}a_n^2$ is absolute convergent too

is a true statement, but before I proof it I ask wether this exercise is probably meant for series in $\mathbb{R}$ respectivly $\mathbb{C}$ or is it meant for ANY series. Like series of vectors in an unknown vectorspace.
Could someone clarify me?
I don't know what I mumble so here is what I stumbled on, while searching for definitions of absolute convergence: https://proofwiki.org/wiki/Definition:Absolute_Convergence/General

Consider a series on a normed vector space $(V,\|\cdot\|)$ where for every $a\in V$ is well defined $a^2$ with $a^2\in V$ and $\|a^2\| = {\|a\|}^2$. The series $\sum a_n = {\left\{ \sum_{n=1}^k a_n \right\}}_{k\ge 1}$ is absolute convergent if
$$
\sum \|a_n\| = \lim_{k\to \infty} \sum_{n=1}^k \|a_n\| < +\infty.
$$
Let us now turn to the statement made. Since the series $\sum a_n$ converge absolute, then only a finite number of terms $a_n$ have the property $ \|a_n\|>1$, otherwise there would be infinite terms $a_n$ with $\|a_n\|>1$, then $\sum_{\|a_n\|>1} 1 = +\infty$ and follows 
$$
\sum \|a_n\| = \sum_{n\colon \|a_n\|>1} \|a_n\| + \sum_{n\colon \|a_n\|\le 1} \|a_n\| > \sum_{n\colon \|a_n\|>1} 1 + \sum_{n\colon \|a_n\|\le 1} 0 \ge +\infty
$$
which is a contradiction. Then, it follows
\begin{align*}
\sum \|a_n^2\| = \sum {\|a_n\|}^2 &= \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum_{n\colon \|a_n\|\le 1} {\|a_n\|}^2\\
&\le \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum_{n\colon \|a_n\|\le 1} \|a_n\| \\
&\le \sum_{n\colon \|a_n\|>1} {\|a_n\|}^2 + \sum \|a_n\| < +\infty,
\end{align*}
since $\sum_{n\colon \|a_n\|>1} {\|a_n\|}^2$ is a finite sum, ${\|a_n\|}^2 \le \|a_n\|$ si $\|a_n\|\le 1$ and $\sum_{n\colon \|a_n\|\le 1} \|a_n\| \le \sum \|a_n\|$. This it, $\sum a_n^2$ is absolute convergent.\
Posdata: I use notation 
$$ 
n\colon \|a_n\|\le 1 = \{n\in \mathbb{N}\colon \|a_n\|\le 1\}.
$$
Also, review this example, if $a_n=\frac{1}{n},~n\ge 1$, then 
$$
\sum a_n^2 < +\infty \quad \text{but} \quad \sum a_n = +\infty.
$$
