I know this question from Abbott's book has been referenced: Prove that if an infinite series converges, the associative property holds . However, I was wondering if the following is an alternative proof for such a statement.
From a theorem in Abbott's book we know any subsequence of a convergent sequence converges to the same limit as the original sequence. Define the following subsequence of the sequence of partial sums on $a_n$: fix $ k$, and $\forall m \in \mathbb{N}$, $ s_{km_1}$, $s_{km_2},... $ (i.e) $s_3, s_6, s_9,...$ . By the theorem above, this converges to the same limit as the original sequence. Now define another subsequence, $s_{k(m_1+1)}, s_{k(m_2+1)}, ...$, i.e the previous sequence shifted by k.
Using these two subsequences define the following sequence: $p_m = s_{k(m_1+1)} - s_{k(m_1)} = a_{km_1+1} + a_{km_1+2}, ... + a_{k(m_1+1)}$ where the $a_n$ are in the original sequence. These $p_m$ represent a grouping of $k$ terms in the original sequence. Define the sequence of partial sums on the $p_m$ including $p_0$, and define $p_0 = s_k$, as $t_1 = p_0 + p_1 = s_{2k} , t_2 = p_0 + p_1 + p_2 = s_{3k}$ thus this sequence of partial sums is a subsequence of the original sequence of partial sums on the $a_n$ and thus converges to the same limit.
I feel like something is definitely missing.
What appears to be missing here is consideration for an arbitrary grouping of sum terms. The associative property would, for example, demand that
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = \left( \frac{1}{1} \right) + \left( -\frac{1}{2} + \frac{1}{3} \right) + \left( -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} \right) + \ldots,$$
which your proof doesn't seem to cover.