Is the Convergence of a Real Number Series Determined by Positive Integers?

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Here is this week's POTW:

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Let $\sum\limits_{n = 1}^\infty a_n$ be a series of real numbers. Show that $\sum\limits_{n = 1}^\infty a_n$ converges absolutely if and only if to each $\epsilon > 0$, there corresponds a positive integer $N = N(\epsilon)$ such that if $n_1,\ldots, n_m$ are indices no less than $N$, then $|a_{n_1} + \cdots + a_{n_m}| < \epsilon$.
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No one answered this week's problem. You can find my solution below.

Spoiler

Suppose $\sum_{n = 1}^\infty a_n$ converges absolutely. Given $\epsilon > 0$, there exists a positive integer $N$ such that $\sum_{n = p}^q |a_n| < \epsilon$ whenever $q \ge p \ge N$. Let $n_1,\ldots, n_k$ be a finite set of indices such that $n_i \ge N$ for all $i$. Without loss of generality, assume $n_1 < \cdots < n_k$. Then

$$|a_{n_1} + \cdots + a_{n_k}| \le |a_{n_1}| + \cdots + |a_{n_k}| \le \sum_{n = n_1}^{n_k} |a_n| < \epsilon.$$

To prove the converse, we argue by contaposition. Suppose that $\sum_{n = 1}^\infty |a_n|$ diverges. Then there exists $\epsilon > 0$ such that, given $N \in \Bbb Z^{+}$, there exists $s > r \ge N$ such that $\sum_{n = r}^s |a_n|\ge \epsilon$. Let $n_1,\ldots, n_m, n_{m+1},\ldots, n_k$ be the indices in $\{r,r+1,\ldots, s\}$ such that $a_n \ge 0$ for $n = n_1,\ldots, n_m$ and $a_n \le 0$ for $n = n_{m+1},\ldots, n_k$. Since $$|a_{n_1} + \cdots + a_{n_m}| + |a_{n_{m+1}} + \cdots + a_{n_k}| = \sum_{n = r}^s |a_i| \ge \epsilon,$$ either $|a_{n_1} + \cdots + a_{n_m}| \ge \epsilon/2$ or $|a_{n_{m+1}} + \cdots + a_{n_k}| \ge \epsilon/2$. Hence, the contrapositive holds.