# Help Understanding Step in Proof of Convergence

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

$$\sum_{i = n}^m |a_i| \le \epsilon.$$

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

$$|s_n - s_n'| \le \epsilon$$ 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.

• What do you mean by first and second inequality? – Snufsan Sep 14 '14 at 8:15

$$\sum_{i = n}^m |a_i| \le \epsilon.$$
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 $\sum_{i=N+1}^q |a_i|$