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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?

Jeremy Jeffrey James
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    $(x_n)$ 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 '19 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 – Felix Mar 14 '19 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 '19 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 '19 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 '19 at 10:57
  • Is the above not in contradiction to this idea here? https://math.stackexchange.com/questions/513073/is-this-valid-every-cauchy-sequence-in-a-normed-space-is-absolutely-convergent?rq=1 – Jeremy Jeffrey James Mar 14 '19 at 11:16
  • Also, suppose $(x_{n})$ converges to 0, then still if $|x_{n}|=\frac{1}{n}|$ the series will not converge absolutely. – Floris Claassens Mar 14 '19 at 12:19
  • Why has this been downvoted? – Jeremy Jeffrey James Mar 14 '19 at 15:59

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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).

Martin R
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