TITLE: Help Understanding Step in Proof of Convergence
QUESTION [5 upvotes]: The theorem is 

If $\sum a_n$ is a series of complex numbers which converges
  absolutely then every rearrangement of $\sum a_n$ converges, and they
  all converge to the same value.

The proof given is as follows:

Let $\sum a_n'$ be a rearrangement, with partial sums $s_n'$. Given
  $\epsilon > 0$, there exists an integer $N$ such that $m \ge n \ge N$
  implies
\begin{equation} \sum_{i = n}^m |a_i| \le \epsilon. \end{equation} 
Now choose $p$ such that the integers $1,2, \cdots, N$ are all
  contained in $k_1, k_2, \cdots, k_p$ where $\{k_n\}$ is a bijective
  function from $\mathbb{N} \rightarrow \mathbb{N}$. Furthermore, $s_n' = s_{k_n}$
    Then if $n>p$, the numbers $a_1, \cdots, a_N$ will cancel in the difference $s_n - s'_n$, so that 
\begin{equation} |s_n - s_n'| \le \epsilon \end{equation} from the
  above inequality. Hence $\{s_n'\}$ converges to the same sum as $\{s_n\}$.

I am having trouble seeing why the second inequality follows from the first. Thanks.

REPLY [2 votes]: Your first inequalty 
\begin{equation} \sum_{i = n}^m |a_i| \le \epsilon. \end{equation} 
says that no matter how many members of the tail of the sum (where the tail in this case, are all members after the $N$-th) you sum up absolutely, their sum will always be smaller than $\epsilon$. 
The difference of sums $|s_n - s_n'|$ contains only elements in the tail of the sequence, finitely many, that means we can find the element with the largest index (from the original sequence $a_n$), lets say its $a_q$ 
(q is the largest index). This means that
$$|s_n - s_n'| \leq \left| \sum_{i=N+1}^{q}a_i \right| \leq \sum_{i = N+1}^q |a_i| \le \epsilon.$$
We bound $|s_n - s_n'|$  with it's maximum value $\left| \sum_{i=N+1}^{q}a_i \right|$, and then use the generalized triangle inequality to bound it with $\sum_{i=N+1}^q |a_i|$