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?
Best Answer
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).