# Abbott's proof that any rearrangement of an absolutely convergent series converges to the same limit as the original

Here is his proof in full:

Assume $$\sum\limits_{k = 1}^{\infty} a_k$$ converges absolutely to $$A$$, and let $$\sum\limits_{k = 1}^{\infty} b_k$$ be a rearrangement of $$\sum\limits_{k = 1}^{\infty} a_k$$. Let's use $$s_n$$ to denote the partial sums of the original series and $$t_m$$ for the partial sums of the rearranged series. Thus, we want to show that $$(t_m) \to A$$.

Let $$\epsilon > 0$$. By hypothesis, $$(s_n) \to A$$, so choose $$N_1$$ such that $$|s_n - A| < \frac{\epsilon}{2}$$ for all $$n \geq N_1$$. Because the convergence is absolute, we can choose $$N_2$$ so that $$\sum\limits_{k = m + 1}^{n} |a_k| < \frac{\epsilon}{2}$$ for all $$n > m \geq N_2$$. Now take $$N = \max \{N_1, N_2\}$$. We know that the finite set of terms $$\{a_1, \ldots, a_N\}$$ must all appear in the rearranged series, and we want to move far enough out in the series $$\sum\limits_{n = 1}^{\infty} b_n$$ so that we have included all of these terms. Thus, choose $$M = \max \{f(k) : 1 \leq k \leq N \}$$.

It should now be evident that if $$m \geq M$$, then $$(t_m - s_N)$$ consists of a finite set of terms, the absolute values of which appear in the tail $$\sum\limits_{k = N + 1}^{\infty} |a_k|$$. Our choise of $$N_2$$ earlier then guarantees $$|t_m - s_N| < \frac{\epsilon}{2}$$, and so $$|t_m - A| = |t_m - s_N + s_N - A| \leq |t_m - s_N| + |s_N - A| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ whenever $$m \geq M$$

I don't understand why the absolute values of $$(t_m - s_N)$$ must appear in the tail of this sequence.

• My bad. I've fixed it – mooglin Jun 26 '20 at 0:43

## 1 Answer

What should be written is "$$t_m - s_N$$ consists of a finite sum of terms, the absolute values of which appear in $$\sum_{k=N+1}^\infty|a_k|$$." This is simply because $$t_m=s_N + \text{other terms a_i where i\notin\{1,\dots,N\}.}$$ This was the point of picking a sufficiently large partial sum of the series $$\sum b_k$$, so that we have included the summands in $$s_N$$.

• Could you also explain what he means by "Our choice of $N_2$ earlier then guarantees $|t_m - s_N| < \frac{\epsilon}{2}"$? – mooglin Jun 26 '20 at 1:32
• @mooglin $t_m-s_N$ is the sum of a finite number of terms $a_i$ where each subscript $i$ exceeds $N$. By triangle inequality, $|t_m-s_N|\le \sum |a_i|$ where the sum is taken over the terms $a_i$ mentioned above. It's a finite number of terms and each $i$ exceeds $N$, which exceeds $N_2$, so ... – grand_chat Jun 26 '20 at 1:51