An equivalent way of saying that a normed vector space is complete is saying that every absolutely convergent series, converges. Hence' in some normed vector-space(incomplete), there must be a absolutely convergent series, that does not converge regularly, but assume that we are in that vector space, and assume that $\Sigma_{n=1}^\infty \|x_n\|$, converges. Why is this proof for showing that the sequence converges regularly then false?
$\|\Sigma_{n=1}^\infty x_n-\Sigma_{n=1}^Kx_n\|=\|\Sigma_{n=K+1}^{\infty}x_n\|$, and by the triangle inequality:
$\le\Sigma_{n=K+1}^{\infty}\|x_n\|$, and this last expression we can get as small as we want by increasing K, because we know that the series converges absolutely.
You are either implicitly assuming the series already converges (to argue that some tail of it has norm as small as you desire), or you are essentially showing that the partial sums form a Cauchy sequence. The latter only suffices to obtain convergence under completeness, which we assumed not to hold.