Proof that every absolutely convergent series is unconditionally convergent

Suppose the series ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$ converges to a $A$ and that it is also absolutely convergent. So, as ${\sum }_{k=1}^{\mathrm{\infty }}|a_k|$ converges, we have $\forall ϵ>0$:

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

Now suppose that ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k$ is a rearrangement of ${\sum }_{k=1}^{\mathrm{\infty }}{a}_k$, that is, it exists a bijection $\sigma :N\to N$, such that $b_k=a_{\sigma (k)}$

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

$$\sum _{k=1}^m{b}_k=b_1+\mathrm{\dots }+b_m=a_{\sigma (1)}+\mathrm{\dots }+a_{\sigma (m)}$$

This sum must include the terms $a_1,\mathrm{\dots },a_n$, otherwise $J$ wouldn’t be maximum. Thus, for $m\ge J$, we have that ${\sum }_{j=1}^m{b}_j-{\sum }_{k=1}^n{a}_k$ is a finite sum of terms $a_k$, with $k\ge n+1$. So:

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

So ${\sum }_{k=1}^{\mathrm{\infty }}{b}_k=A$