When I encountered Banach spaces I was presented with some proofs that link completeness, vector spaces, series and sequences (of partial sums). In particular I was presented with the following theorem:
A normed vector space X is complete if and only if every absolutely summable series in X is summable. That is, X is complete if and only if $\sum_{n=1}^{\infty}x_n$ converges in X whenever $\sum_{n=1}^{\infty}||x_n|| < \infty$.
The problem I am facing is that I can only imagine that every absolutely summable series in X is always summable, as that is what I was taught. (This is what I thought initially; I know now it is not true.)
I then did some searching online, as well as here, and came to the conclusion that this theorem is actually the defining factor in that statement; Completeness links absolutely summable series to summable series, and conversely, when absolutely summable series are summable they define a complete space. This must mean that all 'examples' I can think of (or in other words, the scope of my imagination) are examples in Banach spaces.
This however raises the following question:
Can anyone provide me with some examples of absolutely summable series in a normed vector space that are not summable, and explain why?
To clarify further, my goal is to understand what is the defining difference between being absolutely summable and summable. I suppose that the norm (or metric) used is key, as this defines when $\sum_{n=1}^{\infty}||x_n|| < \infty$. I just don't see it yet. If there is any way to link the explanation to Cauchy sequences, as most proofs use that approach, that would be great. Thanks!
Best Answer
Take $\mathbb Q$, which is a vector space over $\mathbb Q$, and let $0.a_1a_2\ldots$ be the decimal expansion of $\sqrt2-1$. Then $\sum_{n\geq1}a_n10^{-n}$ does not converge in $\mathbb Q$, but we have that $\sum_{n\geq1}\|a_n10^{-n}\|=\sum_{n\geq1}a_n10^{-n}$ converges (to $\sqrt2-1$, of course).
For a less trivial example, take $C_c(\mathbb R)\subset L^2(\mathbb R)$, approach $\exp(-x^2)\in L^2$ by a sequence $(u_n)$ of continuous functions with compact supports, rewrite this as a series $\sum_{n\geq1}f_n$ with $f_n=u_n-u_{n-1}$ (agreeing on $u_0=0$). Then the series is absolutely summable, but not summable, because the limit $\exp(-x^2)$ is not in $C_c(\mathbb R)$.
The point is that an absolutely convergent series need not converge since in a non-complete space a Cauchy sequence need not converge. Each non-complete space can be used to construct such an example. Note that an absolutely summable series defines a Cauchy sequence (of its partial sums). Hence it should converge in the completion $\overline X$ of your original space $X$. The only way that the sequence does not converge in $X$, is when $X\neq \overline X$, so when $X$ is not complete.