Convergence of series in inclomplete normed vector space I tried to prove that in non-Banach normed vector space there always exists a series which converges absolutely but which does not converge.
The idea was to consider a Cauchy sequence that doesn't converges and try to construct series with that property, but obvious methods like considering $x_n - x_{n-1}$ didn't work. 
So, any hints?
 A: Let $\{ x_n \}_{n=1}^{\infty}$ be a Cauchy sequence that does not converge. If any subsequence of a Cauchy sequence converges to $x$, then the original Cauchy sequence converges to $x$ as well. Therefore, no subsequence $\{ x_{n_k} \}_{k=1}^{\infty}$ of $\{ x_n \}_{n=1}^{\infty}$ converges. Choose
$$
           n_1 < n_2 < n_3 < \cdots
$$
such that
$$
                 \|x_{n}-x_{m}\| < \frac{1}{2^k},\;\;\; n,m \ge n_k.
$$
Then $\|x_{n_i}-x_{n_j}\| < \frac{1}{2^k}$ for $i,j\ge k$. Therefore, the following cannot converge, even though the sum on the right is absolutely convergent:
\begin{align}
         x_{n_k} &= x_{n_1}+(x_{n_2}-x_{n_1})+(x_{n_3}-x_{n_2})+\cdots+(x_{n_k}-x_{n_{k-1}}) \\ &= x_{n_1} +\sum_{j=1}^{k}(x_{n_{j}}-x_{n_{j-1}}).
\end{align}
A: Using the proof of TrialAndError one sees that 
Proposition In every normed space (NS) the following properties are equivalent
a) the space is Banach 
b) every absolutely (for an arbitrary NS, one should say "normally") convergent series is convergent. 
