Does the series corresponding to a Cauchy sequence **always** converge absolutely?

Let $$X$$ be a normed vector space and consider a Cauchy sequence $$(x_n)_{n\in\mathbb{N}}$$ in $$X$$.

Is it true that the corresponding series of our Cauchy sequence, $$\sum_{i=1}^\infty x_i$$, always converges absolutely? (that is to say $$\sum_{i=1}^\infty \|x_i\|_X$$ converges)

If not, what are some counter examples to elucidate the point?

• $(x_n)$ defined by $x_n = 13$ is a Cauchy sequence in $\Bbb R$, but $\sum x_n$ does not converge at all. – Martin R Mar 14 at 10:34
• additionally to Martin R's comment, maybe to do anything with cauchyness and convergence, you maybe want to assume that your space in complete – Enkidu Mar 14 at 10:36
• I understand that in a complete NVS every Cauchy sequence must converge (by definition). I was wondering whether in a purely NVS the series corresponding to a Cauchy sequence was guaranteed to converge absolutely. – Jeremy Jeffrey James Mar 14 at 10:38
• @JeremyJeffreyJames: It seems that you are mixing up the sequence and the series. A convergent sequence $(x_n)$ is a Cauchy sequence, but that does not imply the convergence of of the series $\sum x_n$ at all, neither absolutely nor conditionally. – Martin R Mar 14 at 10:53
• This relates to another question - which I will now ask and then link to this one - @MartinR, if you post your counter example as an answer I will happily accept. – Jeremy Jeffrey James Mar 14 at 10:57

For a sequence $$(x_n)$$ we have the following implications: $$\begin{matrix} (x_n)_n \text{ convergent} & \implies & (x_n)_n \text{ Cauchy} \\ \Downarrow & & \Downarrow \\ (\Vert x_n \Vert )_n \text{ convergent} & \implies & (\Vert x_n \Vert )_n \text{ Cauchy} \\ \end{matrix}$$ (and the “horizontal” implications are equivalences if $$X$$ is complete).
But none of this implies that the corresponding series $$\sum x_n$$ converges at all, a simple counter-example is a constant non-zero sequence.
The series $$\sum x_n$$ is convergent if the sequence $$(s_n)_n$$ of partial sums $$s_n = \sum_{k=1}^n x_k$$ is convergent. With $$S_n = \sum_{k=1}^n \Vert x_k \Vert$$ we have $$\begin{matrix} \sum \Vert x_n \Vert \text{ convergent} & \iff (S_n)_n \text{ convergent} & \implies & (S_n)_n \text{ Cauchy} \\ & & & \Downarrow \\ \sum x_n \text{ convergent} & \iff (s_n)_n \text{ convergent} & \implies & (s_n)_n \text{ Cauchy} \\ \Downarrow \\ \lim_{n\to \infty} x_n = 0 \end{matrix}$$
If $$X$$ is complete then the horizontal implications are equivalences, to that $$\sum \Vert x_n \Vert \text{ convergent} \implies \sum x_n \text{ convergent} \, .$$
But again, $$(x_n)$$ being Cauchy (or convergent) does not imply that $$\sum x_n$$ is convergent (conditionally or absolutely).