FALSE: if some grouping of a series converges then the series itself converges

False claim: if $(n_j)$ is strictly increasing with $n_0 = 0$ and the series of blocks $\sum_j B_j$, $B_j = \sum_{k=n_j}^{n_{j+1}-1} a_k$, converges (Series, partial sums, convergence and the sum, divergence, and the tail series), then $\sum a_k$ converges.

What is true is the opposite direction, 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: convergence of $\sum a_k$ implies convergence of every grouping, to the same sum. Brackets may be inserted into a convergent series; they may not be removed.

The witness is the alternating sequence itself. 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 and odd index maps $e$ and $o$ satisfying $o_j = e_j + 1$ and $e_{j+1} = o_j + 1$, and group in pairs, $n_j := e_j$. Every block is $\varepsilon_{e_j} + \varepsilon_{o_j} = 1 + (-1) = 0$, so the grouped series is $0 + 0 + \dots$ and converges to $0$; but $\sum \varepsilon_k$ diverges, its terms having absolute value $1$ and so not tending to $0$ (If a series converges then its terms tend to $0$).

• The blocks hide a cancellation the grouped series cannot see. Each block is the sum of two terms of absolute value $1$; grouping reports only their sum, and the information that destroys convergence lives strictly inside a block. This is why 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 never looks inside a block and why its converse is hopeless in general.

• What the witness does and does not isolate. Its block lengths are all equal to $2$, so no amount of control on the block lengths alone repairs the claim. What it does violate is $a_k \to 0$, which by If a series converges then its terms tend to $0$ is necessary for convergence of $\sum a_k$ and is invisible to the grouped series. Whether adding that condition to the hypothesis repairs the claim is not decided here and is not needed anywhere on this page.

• The same series is the classical Grandi series. Grouped as $(1-1)+(1-1)+\dots$ it appears to sum to $0$, and grouped as $1 + (-1+1) + (-1+1) + \dots$ it appears to sum to $1$. Both are groupings in the sense 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: the second is $n = (0,1,3,5,\dots)$, strictly increasing with $n_0 = 0$, whose first block is the single term $a_0$. So the two values are genuinely produced by two admissible groupings, and no hypothesis of that theorem excludes either; what fails is its hypothesis on the original series, which does not converge, and that is precisely why it says nothing about the grouped sums here. The same pair appears as a counterexample on the companion examples page.