Absolutely convergent series has constant sum Show that the sum of an absolutely convergent series does not change if the terms are rearranged.
Let the absolutely convergent series be $a_1,a_2,\ldots$, and let the rearrangement be $b_1,b_2,\ldots$. By the Cauchy criterion of convergence, since the series $|a_1|,|a_2|,\ldots$ converges, it follows that for any $\epsilon$, there exists $N$ such that $|a_m|+|a_{m+1}|+\ldots+|a_n|<\epsilon$ for all $n>m>N$. 
We'll show that the series $b_i$ converges. Let $N'$ be the greatest index such that $b_{N'}$ comes from one of $a_1,a_2,\ldots,a_N$. Then for all $n>m>N'$, we have $|b_m+b_{m+1}+\ldots+b_n|\leq |b_m|+|b_{m+1}|+\ldots+|b_n|<\epsilon$.
So the series $b_i$ converges. How can we show it converges to the same sum as $a_i$?

Say w.l.o.g. $\sum a_n=c_1<c_2=\sum b_n$. Take $0<\epsilon <\frac{c_2-c_1}{3}$. Then take $N>0$ such that for all $m>N$ we have
$$\left|\sum_{n\leq m} b_n - c_2\right|<\epsilon; \quad \sum_{n>N}|b_n|<\epsilon$$
and take $M>0$ such that $\{a_1,\ldots,a_M\}$ contain $\{b_1,\ldots,b_N\}$ and
$$\left|\sum_{n\leq M} a_n - c_1\right|<\epsilon.$$
Then
$$c_2-c_1< (\sum_{n\leq N} b_n + \epsilon) - (\sum_{n\leq M} a_n - \epsilon)\leq 2\epsilon + \sum_{n>N}|b_n| < 3\epsilon< c_2-c_1.$$
