# 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 ProofThatEveryAbsolutelyConvergentSeriesIsUnconditionallyConvergent 2013-03-11 19:17:17 2013-03-11 19:17:17 Filipe (28191) (0) 7 Filipe (0) Proof msc 40A05