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.
Best 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 $s_N$.