Every rearrangement of an absolutely convergent series converges to the same sum (Rudin)

Question

What guarantees $\sum_{i=n}^m |a_i|$ will not be less than $|s_n-s_n'|$; hence $|s_n-s_n'| \ge \epsilon$?

May someone explain, please?

Answer 1

I don't quite follow the notation, but I see what he's saying. Since the series converges, we know that for every $\epsilon > 0$ there exists a $N = N(\epsilon)$ such that $\sum_{N+1}^{\infty}{|a_n|} < \epsilon$. Now $N$ is finite, so if we go out far enough in the rearrangement, say to M, we will encounter all the summands $a_1, a_2,..., a_n$ among the summands $a_1', a_2',...a_M'.$ Now look at $|s_M-s_M'|.$ All the terms up to $a-N$ have been cancelled out, by construction of M. Some other terms are cancelled out also, but what matters is that any terms remaining are of the $|a_k|$ with $k>N$. So, $|s_M-s_M'| <= \sum_{N+1}^{\infty}{|a_n|} < \epsilon.$