$(1-1)+(1-1)+\dots$ 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 $({\epsilon }_k)$ 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 $\mathrm{\mid }{s}_k\mathrm{\mid }=1$, $s\circ e\equiv 1$ 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 ${\sum }_k{\epsilon }_k$ in consecutive pairs, $n_j:=e_j$. Each block is

$B_j={\epsilon }_{e_j}+{\epsilon }_{o_j}=1+(-1)=0,$

so the grouped series is $0+0+0+\dots$ and converges to $0$, while ${\sum }_k{\epsilon }_k$ 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 $\sum {\epsilon }_k$ 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 $2$.