Proof that every absolutely convergent series is unconditionally convergent

Suppose the series $\sum_{k=1}^{\infty}a_{k}$ converges to a $A$ and that it is also absolutely convergent. So, as $\sum_{k=1}^{\infty}|a_{k}|$ converges, we have $\forall\epsilon>0$ :

$\exists N>0:\sum_{k=n+1}^{\infty}|a_{k}|<\frac{\epsilon}{2},\forall n>N$

As the series converges to $A$ , we may choose $n$ such that:

$|A-\sum_{k=1}^{n}a_{k}|<\frac{\epsilon}{2}$

Now suppose that $\sum_{k=1}^{\infty}b_{k}$ is a rearrangement of $\sum_{k=1}^{\infty}a_{k}$ , that is, it exists a bijection $\sigma:\mathbb{N}\rightarrow\mathbb{N}$ , such that $b_{k}=a_{\sigma(k)}$

Let $J=max\{j:\sigma(j)\leq n\}$ . Now take $m\geq J$ . Then:

$\sum_{k=1}^{m}b_{k}=b_{1}+...+b_{m}=a_{\sigma(1)}+...+a_{\sigma(m)}$

This sum must include the terms $a_{1},...,a_{n}$ , otherwise $J$ wouldn’t be maximum. Thus, for $m\geq J$ , we have that $\sum_{j=1}^{m}b_{j}-\sum_{k=1}^{n}a_{k}$ is a finite sum of terms $a_{k}$ , with $k\geq n+1$ . So:

$|\sum_{j=1}^{m}b_{j}-\sum_{k=1}^{n}a_{k}|\leq\sum_{k=n+1}^{\infty}|a_{k}|<% \frac{\epsilon}{2}$

Finally, taking $m\geq J$ , we have:

$|A-\sum_{k=1}^{m}b_{k}|\leq|A-\sum_{k=1}^{n}a_{k}|+|\sum_{j=1}^{m}b_{j}-\sum_{% k=1}^{n}a_{k}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$

So $\sum_{k=1}^{\infty}b_{k}=A$

Title	Proof that every absolutely convergent series is unconditionally convergent
Canonical name	ProofThatEveryAbsolutelyConvergentSeriesIsUnconditionallyConvergent
Date of creation	2013-03-11 19:17:17
Last modified on	2013-03-11 19:17:17
Owner	Filipe (28191)
Last modified by	(0)
Numerical id	7
Author	Filipe (0)
Entry type	Proof
Classification	msc 40A05
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