Grouping Terms in Absolutely Convergent Series

Why Grouping Needs Care

A series is defined through the limit of its partial sums. Grouping terms changes which partial sums are displayed, but it need not change their order. This distinction matters: a valid grouping combines finite runs of consecutive terms, and its partial sums are then selected partial sums of the original series.

For an absolutely convergent series, grouping consecutive terms preserves the sum and leaves the grouped series absolutely convergent. In fact, the preservation of the sum requires less: any convergent series keeps its sum under such a grouping. A conditionally convergent series gives a useful comparison, because grouping consecutive terms still works even though absolute convergence fails.

What Counts as Grouping Consecutive Terms?

Let \(\sum_{n=1}^{\infty} a_n\) be a series. To group its terms into consecutive blocks, choose integers \(0=n_0<n_1<n_2<\cdots\) with \(n_k\to\infty\). The \(k\)th block consists of the terms with indices \(n_{k-1}+1,\ldots,n_k\). Each block is finite, the blocks do not overlap, and together they include every term.

For example, the blocks \((a_1+a_2),(a_3+a_4+a_5),(a_6)\) are consecutive. A block containing \(a_1\) and \(a_3\) but omitting \(a_2\) is not. Nor is a single “block” consisting of infinitely many terms allowed here.

The Grouping Theorem

The first conclusion is about the sum. The second is an additional guarantee provided by absolute convergence: not only does the grouped series converge to the same value, but the sum of the absolute values of its grouped terms is finite.

Why the Theorem Works

Write \(S_N=\sum_{n=1}^{N}a_n\) for the \(N\)th partial sum of the original series. After \(m\) grouped terms have been added, every term through index \(n_m\) has been included exactly once. Thus the \(m\)th grouped partial sum \(T_m\) is exactly \(S_{n_m}\).

This proves that the grouped series has the same sum. Notice that the argument used convergence, but not absolute convergence. The absolute convergence part follows separately: the triangle inequality for each finite block gives

Summing this inequality over the first \(m\) blocks gives

If \(\sum |a_n|\) is finite, the partial sums of \(\sum |b_k|\) are increasing and bounded above by that finite total. They therefore converge, so \(\sum |b_k|\) converges. This is precisely absolute convergence of the grouped series.

Worked Examples

Comparison with a Conditionally Convergent Series

The alternating harmonic series \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\) converges by the alternating-series test: its positive term magnitudes decrease to zero. It is not absolutely convergent, because its magnitude series is the divergent harmonic series. Nevertheless, grouping consecutive terms into pairs preserves its sum.

This example sharpens the role of absolute convergence. It guarantees that the grouped series is absolutely convergent, but it is not necessary for preserving the sum under finite consecutive grouping. Convergence of the original series is enough for that conclusion.

A Common Pitfall: Treating Infinite Collections as Blocks

A tempting but invalid operation on the alternating harmonic series is to collect all its positive terms into one group and all its negative terms into another, then subtract the two “group sums.” The positive terms have sum \(\sum_{k=1}^{\infty}1/(2k-1)=+\infty\), while the magnitudes of the negative terms have sum \(\sum_{k=1}^{\infty}1/(2k)=+\infty\). Writing \(+\infty-\infty\) and assigning it a finite value is not valid arithmetic.

The practical test is straightforward: check that every block has finitely many terms, that the blocks are consecutive and disjoint, and that their endpoints tend to infinity. Then each grouped partial sum is an original partial sum, which is the essential reason the sum is preserved.