Absolute convergence does not imply convergence Find a space and a series that converges absolutely but it does not converges.
It is clear that the space can't complete or Banach.
 A: You can take $X$ to be the vector space of polynomials on the unit circle $\mathbb T$, with the supremum norm. Now consider the series
$$
\sum_{n=1}^\infty \frac{x^n}{2^n}.
$$
The series converges absolutely, because $|x|=1$, and you get a numerical series that converges in $\mathbb C$. But the series cannot converge in $X$, because it would have to be $\frac x {2-x}$, which is not a polynomial.
A: The taylor series expansion of $\sqrt{1+x}$ is 
$$1+a_1x-a_2x^2+a_3x^3+\cdots,$$
or
$$1+\sum_1^\infty (-1)^{n+1}a_nx^n,$$
for $|x|\le1$, and $a_1, a_2, a_3, \cdots$ are all positive rational numbers.
Now, by simply substituting $x=-1$ we get
$$1-\sum_1^\infty a_n = \sqrt{1 + (-1)}=0,$$
or
$$\boxed{\sum_1^\infty |(-1)^{n+1}a_n| = 1}$$
On the other hand, if you substitute $x=1$ you get
$$\boxed{\sum_1^\infty (-1)^{n+1}a_n = \sqrt{2}-1}$$
which is not convergent in $\mathbb{Q},$ although it's absolutely convergent.
A: Or take the finite sequences $l^2_0$ within the square integrable ones $l^2$ and consider the sequence (or the net):
$$\sum_{k=1}^n\frac{1}{k}e_k$$
$$\sum_{n\in\mathbb{N}}\frac{1}{n}e_n$$
