$(1-1) + (1-1) + \dots$ converges to $0$ while $\sum_{k} (-1)^{k}$ diverges

Refuted claim: if some grouping of a series converges, so does the series (Grouping: if $\sum a_k$ converges and $(n_j)$ is strictly increasing with $n_0 = 0$, the series of blocks $\sum_{k=n_j}^{n_{j+1}-1} a_k$ converges to the same sum, Series, partial sums, convergence and the sum, divergence, and the tail series).

Let $(\varepsilon_k)$ be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$), with even index map $e_j$ satisfying $e_0 = 0$ and $e_{j+1} = e_j + 2$, and group the series $\sum_k \varepsilon_k$ in consecutive pairs, $n_j := e_j$. Each block is

$B_j \;=\; \varepsilon_{e_j} + \varepsilon_{o_j} \;=\; 1 + (-1) \;=\; 0 ,$

so the grouped series is $0 + 0 + 0 + \dots$ and converges to $0$, while $\sum_k \varepsilon_k$ diverges, its terms having absolute value $1$ and so not tending to $0$ (If a series converges then its terms tend to $0$). This is FALSE: if some grouping of a series converges then the series itself converges exhibited.

The partial sums of $\sum \varepsilon_k$ are $0, 1, 0, 1, \dots$, and the grouping picks out exactly the even-indexed ones, all equal to $0$. That is what Grouping: if $\sum a_k$ converges and $(n_j)$ is strictly increasing with $n_0 = 0$, the series of blocks $\sum_{k=n_j}^{n_{j+1}-1} a_k$ converges to the same sum says a grouping always does: it reads a subsequence of the partial sums. A subsequence of a divergent sequence may of course converge, which is the whole of the phenomenon.

• A second bracketing gives a different answer, and it is just as legitimate. Grouping as $1 + (-1+1) + (-1+1) + \dots$ is the grouping $n = (0, 1, 3, 5, \dots)$, whose blocks are $\{0\}$ and then the pairs $\{2j-1, 2j\}$ for $j \ge 1$. That $n$ is strictly increasing with $n_0 = 0$, so it satisfies the hypotheses of Grouping: if $\sum a_k$ converges and $(n_j)$ is strictly increasing with $n_0 = 0$, the series of blocks $\sum_{k=n_j}^{n_{j+1}-1} a_k$ converges to the same sum exactly as the pairing used above does; its grouped partial sums are $A_{n_m}$, that is $0, 1, 1, 1, \dots$, so this grouping converges to $1$ while the pairing from index $0$ converges to $0$. Two admissible groupings of one series with two different sums is the classical paradox attached to $\sum(-1)^k$, and nothing in Grouping: if $\sum a_k$ converges and $(n_j)$ is strictly increasing with $n_0 = 0$, the series of blocks $\sum_{k=n_j}^{n_{j+1}-1} a_k$ converges to the same sum is violated: that theorem says what the grouped sums are only when the original series converges, and here it does not.

• No sum is being assigned to a divergent series here. Both statements are about ordinary convergence in the sense of Series, partial sums, convergence and the sum, divergence, and the tail series: the grouped series converges, the original does not. Nothing on this page attaches a value to $\sum(-1)^k$.

• What would have to be added. By If a series converges then its terms tend to $0$, $a_k \to 0$ is necessary for the original series to converge and is invisible to the grouped series; this witness violates exactly that condition, with blocks of the constant length $2$.