Series in incomplete normed space We have known that "A normed space $X$ is a Banach space if and only if each absolutely convergent series in X converges". We would like to find an explicitly incomplete normed space and an explicitly  series in that space such that the given series is absolutely convergent but not convergent.
 A: This seems to me to be a relatively simple example:
$X=$ set of all real sequences with finite support (i.e., there are only finitely many non-zero elements)
$\|x\|=\sup\limits_{n\in\mathbb N} |x_n|$
Consider the sequence $a_n=(0,\dots,0,\frac1{n^2},0,0,\dots)$ and the series $\sum a_n$ in $X$.
This series is absolutely convergent, since $\sum \|a_n\|= \sum\frac1{n^2}$.
It cannot be convergent in $X$. Take any sequence $x$ with finite support. This means that there is $n_0$ such that $x_n=0$ for each $n\ge n_0$.
If $s_n=\sum\limits_{k=1}^n a_k$ denotes the $n$-th partial sum, we have 
$$\|s_n-x\| \ge \frac1{n_0^2}$$
for each $n\ge n_0$. So w have $\|s_n-x\|\not\to0$ and $s_n\not\to x$.
A: Following the hint of Qiaochu Yuan i found the solution for my question. The construction of a series in an incomplete normed space that is absolutely convergent but not convergent follows from the proof of the theorem "A normed space X is a Banach space if and only if each absolutely convergent series in X converges".
Let $X$ be an incomplete normed space and $\{x_n\}\subset X$ is a Cauchy sequence that is not convergent. For every $k\geq 1$ there exists $n_k\geq k$ such that
$$
\|x_p-x_q\|<\frac{1}{2^k} \quad \forall p,q\geq n_k.
$$
Without of loss generality we can assume that
$$
n_1<n_2<\ldots<n_k<\ldots
$$
Let $y_k=x_{n_{k+1}}-x_{n_k}$. Then
$$
\sum_{k=1}^{\infty}\|y_k\|=\sum_{k=1}^{\infty}\|x_{n_{k+1}}-x_{n_k}\|\leq\sum_{k=1}^{\infty}\frac{1}{2^k}<+\infty.
$$
Hence $\sum_{k=1}^{\infty}\|y_k\|$ is convergent. We observe that
$$
\sum_{k=1}^{m}\|y_k\|=x_{n_{m+1}}-x_{n_1}
$$
We conclude that $\displaystyle\sum_{k=1}^{\infty}y_k$ is not convergent. If this sequence is convergent then $\{x_{n_m}\}$ is convergent. In addition $\{x_n\}$ is a Cauchy sequence. Then $\{x_n\}$ is convergent which is an absurd.
Note. We can choose $X=C^{L}_{[0,1]}$, the space of continuous functions on $[0,1]$ with the norm given by
$$
\|x\|_{X}=\int_0^1|x(t)|dt.
$$
Let $\{x_n(t)\}_{n\in \mathbb{N}}\subset X$ be given by
$$
x_n(t)=\begin{cases}
1& 0\leq t\leq \frac{1}{2}\\
0& \frac{1}{2}+\frac{1}{2n}\leq t\leq 1\\
n+1-2nt& \frac{1}{2}\leq t \leq \frac{1}{2}+\frac{1}{2n}.
\end{cases}
$$
