Weierstrass M-test, Banach spaces, and absolute convergence As per the Weiestrass M-test, if absolute convergence of the series associated with a sequence in some normed vector space $X$ implies convergence of the series, then $X$ is Banach. However, I think it can be proved that if a series converges absolutely then it necessarily converges. So what is the point I'm missing here?
 A: Consider $c_{00}$, the space of all sequences that are eventually zero. Put the supremum norm on $c_{00}$ and let $(a_n)_n$ be a sequence in $c_{00}$ such that 
$$
a_n(m) =\left\{ \begin{matrix} \frac{1}{n!} & \text{ if } m =n \\ 0 & \text{ else } \end{matrix} \right.
$$
Clearly, for every $n \in \mathbb{N}$, $a_n \in c_{00}$. Also note that
$$
\sum_{n=1}^\infty \| a_n\|_{\infty} = \sum_{n=1}^\infty \frac{1}{n!} = e,
$$
so the series $\sum_{n=1}^\infty a_n$ clearly converges absolutely, but it won't converge in $c_{00}$.
A: For a general counterexample take a non-Banach space $X$.
Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $X$ which is not convergent.
There exists a subsequence $(x_{p(n)})_{n=1}^\infty$ such that $\|x_{p(n+1)} - x_{p(n)}\| = \frac{1}{2^n}$, for $n \in \mathbb{N}$.
Now define a sequence $(y_n)_{n=1}^\infty$ as
$$y_1 = x_{p(1)}$$
$$y_n = x_{p(n+1)} - x_{p(n)}, \text{ for } n \ge 2$$
and consider the series $\sum_{n=1}^\infty y_n$.
It certainly converges absolutely since:
$$\sum_{n=1}^\infty \|y_n\| = \|x_{p(1)}\| + \sum_{n=2}^\infty \|x_{p(n+1)} - x_{p(n)}\| \le \|x_{p(1)}\| + \sum_{n=2}^\infty \frac1{2^n} = \|x_{p(1)}\| + \frac12 < +\infty$$
However, it does not converge:
$$\sum_{n=1}^N y_n = x_{p(N+1)} - x_{p(1)}$$
This converges when $N \to \infty$ if and only if the sequence $(x_{p(n)})_{n=1}^\infty$ converges, which cannot be true since then the entire sequence $(x_{n})_{n=1}^\infty$ would converge.
