Non-example for total convergence theorem. Let's consider the following result: "Let $(X, || \cdot ||)$ be a Banach space. Let $(a_n)_{n\in \mathbb N} \subset X$ be a succession such that the series $\displaystyle \sum_{n\in \mathbb N} ||a_n||$ converges. Therefore $\displaystyle \sum_{n\in \mathbb N} a_n$ converges." 
Now, I wonder what happens if we remove the Banach hypothesis: may you show me an example of a normed vector space (non-Banach) $(X, ||\cdot||)$ such that a succession converges totally but it does not converge in $(X, ||\cdot||)$? Thank you.
 A: Let $A:=\{(x_1, x_2, \dots) \mid \mathrm{Only \, finitely \, many \, nonzero}\}$ equipped with the $\|x\|:=\sup_{i \in \mathbb N}|x_i|$.
Consider the sequence $x_n:=\frac{1}{2^n}\cdot(1,1/2,1/4,1/8, \dots)$.
Then, $\sum_{n \in \mathbb N}\|x_n\|:=\sum_{n \in \mathbb N} \frac{1}{2^n}=1$.
On the other hand, taking a sum over all $x_n$, we see that we in fact get
$$(2,1,1/2, \dots) \notin A.$$
The point is that this absolute convergence condition is exactly equivalent to being Banach. In particular, if you assume that a space is not Banach, then there is a Cauchy sequence $(x_n)$ so that it does not converge. From this, we know that for all $i \in \mathbb N$ there exists some $N_i$ so that $m,n>N_i$ implies that $\|x_n-x_m\| \leq \frac{1}{2^i}$. From which we can consider
$$\sum_{n \in \mathbb N} \|x_{N_{i+1}}-x_{N_i}\|$$
which converges, while $y_n:=x_{N_{i+1}}-x_{N_i}$  probably will not.
From this, we gather that there should be a bunch of counterexamples whenever a space is not Banach.
A: If you take $X=C_{00}$ with sup norm and look at the series $$(1,0,\ldots,0,\ldots)+\left(0,\frac{1}{2^2},\ldots,0,\ldots\right)+\ldots+\left(0,0,\ldots,0,\frac{1}{n^2},0,\ldots \right)+\ldots=\sum_
{n=1}^{\infty}\frac{1}{n^2}e_n$$ where $e_n$ is the usual orthonormal set. Then you can see that this series converges to $\left(1,\frac{1}{2^2},\ldots,\frac{1}{n^2},\ldots\right)$ which is in $C_0$ but not $C_{00}$.
