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Let $V$ be a normed space with norm $\|\|$. $\sum_{n=1}^\infty a_n$ is an absolutely convergent series if $\sum_{n=1}^\infty \|a_n\|$ converges.

Could you please explain why in complete normed spaces the absolute convergence implies convergence, but it doesn't hold for incomplete normed spaces?

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    $\begingroup$ If your space is incomplete, then you can find a Cauchy sequence $s_n$ which is not convergent. Now consider $a_n = s_n-s_{n-1}$. Then $\sum_{k=1}^n a_k=s_n$ is Cauchy (hence absolutely convergent) but not convergent. $\endgroup$
    – Crostul
    Mar 10, 2017 at 8:59
  • $\begingroup$ Useful link: math.stackexchange.com/questions/1583504/… $\endgroup$
    – Konstantin
    Mar 10, 2017 at 14:28

3 Answers 3

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This is the proof of the fact that, if $V$ is a complete normed space, then every absolutely convergent series is convergent. You will see that we need through the proof the fact that $V$ is complete.

Let $\sum_{n=1}^\infty \lVert a_n\rVert$ be a convergent series, which means that $\sum_{n=1}^\infty a_n$ is absolutely convergent. We want to prove that this implies that $\sum_{n=1}^\infty a_n$ is a convergent series. But by definition of a convergent series, it will be convergent if and only if the sequence $\{S_k\}_{k=1}^\infty=\{\sum_{n=1}^k a_n\}_{k=1}^\infty$ is convergent. Since $V$ is complete, this sequence will be convergent if and only if it is a Cauchy sequence. (What we can prove is that this will always be a Cauchy sequence, but if the space $V$ was not complete, we couldn't conclude from here that it would be convergent).

And indeed, it is a Cauchy sequence because:

$$\lVert S_k-S_p\rVert=\biggl\lVert\sum_{n=p+1}^k a_n\biggr\rVert\leq\sum_{n=p+1}^k \lVert a_n\rVert$$

and this expression tends to zero when $k$ and $p$ are taken as close as necessary. This way we have proven that the series $\sum_{n=1}^\infty a_n$ is convergent, as we needed to show.

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    $\begingroup$ You should say "sequence" instead of "succession". $\endgroup$
    – Crostul
    Mar 10, 2017 at 22:18
  • $\begingroup$ @Crostul Oh, sorry, I translated it too literally from Spanish. Thank you! I've edited it. $\endgroup$ Mar 10, 2017 at 22:19
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It's because in an incomplete space the presumed limit does not need to exist within the space. This can happen even if the sum converges absolutely to a value within the space.

To construct an example consider a number $a$ with decimal expansion with $1$ and $3$ alternating in non-periodic manner. Now we can construct a series that converges to this number by alternately putting $5\times10^{-n}-4\times10^{-n}$ and $n$ and $6\times10^{-n}-3\times10^{-n}$. On sees that $a\notin\mathbb Q$, but the sum converges to $a$ in $\mathbb R$. We also see that the absolute series converges (to $1$) in $\mathbb Q$ despite the series does not converge in $\mathbb Q$.

In a complete space on the other hand we have only to prove that absolute convergence means that the sub sums form a cauchy sequence which would mean that it's convergent. But the difference of two sub sums are:

$$|\sum_0^k a_j - \sum_0^l a_n| = |\sum_k^l a_j| \le \sum_k^l |a_j|$$

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complete space: any Cauchy sequence converges.

$A_n:=\sum_{k=1}^na_k$. Then $\|A_n-A_m\|\leq \sum_{k=n}^m\|a_k\|$. Since $\sum_k \|a_k\|$ converges, $\{A_n\}$ is a Cauchy sequence. Now the completeness of the space implies the convergence of $A_n$. Without completeness, we cannot claim the convergence of $A_n$.

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