Rearrangements and Riemann's Rearrangement Theorem

Technique: $\varepsilon$-tail estimate using absolute convergence, via the Cauchy criterion. Let $S = \sum_{n=1}^{\infty} a_n$ and let $\varepsilon > 0$ be given.

Step 1 (tail smallness). Since $\sum |a_n| < \infty$, the sequence of partial sums of $\sum|a_n|$ is Cauchy, so there exists $N \in \mathbb{N}$ such that

(This sum is finite because $\sum |a_n|$ converges, so its tail past any point is a nonnegative convergent series and shrinks to $0$.) Consequently, if $S_N = \sum_{n=1}^N a_n$ denotes the $N$-th partial sum of the original series, then

using the triangle inequality for the (finite-tail) sum, extended to the infinite tail by taking a limit of finite partial tails, which is legitimate since the tail series converges absolutely.

Step 2 (finding where all the first $N$ terms reappear). Because $\sigma$ is a bijection, each of $1, 2, \ldots, N$ appears exactly once among $\sigma(1), \sigma(2), \ldots$. Let

This $M$ is a well-defined natural number since it is the maximum of a finite set of natural numbers. By construction, for every $m \ge M$, the finite set $\{\sigma(1), \ldots, \sigma(m)\}$ contains $\{1, 2, \ldots, N\}$ as a subset (every index $k \le N$ has $\sigma^{-1}(k) \le M \le m$, so $k = \sigma(\sigma^{-1}(k))$ appears among the first $m$ rearranged terms).

Step 3 (comparing partial sums of the rearrangement to $S_N$). Fix any $m \ge M$, and let $T_m = \sum_{k=1}^m a_{\sigma(k)}$. By Step 2, $\{1, \ldots, N\} \subseteq \{\sigma(1), \ldots, \sigma(m)\}$, so

because the terms indexed by $1, \ldots, N$ appear in both sums and cancel exactly (each exactly once, since $\sigma$ is injective on $\{1,\ldots,m\}$), leaving only rearranged terms whose original index exceeds $N$. Every index $j$ appearing in this remaining sum satisfies $j > N$ and $j \in \{\sigma(1), \ldots, \sigma(m)\}$, a finite set of distinct indices all exceeding $N$. Hence, by the triangle inequality and Step 1's tail estimate,

(The sum on the right is over all indices $j > N$, which only enlarges a sum of nonnegative terms, so the inequality is valid.)

Step 4 (combining). For all $m \ge M$, by (1) and (2) and the triangle inequality,

Since $\varepsilon > 0$ was arbitrary, this shows $T_m \to S$ as $m \to \infty$, i.e. $\sum_{n=1}^\infty a_{\sigma(n)} = S = \sum_{n=1}^\infty a_n$.

Step 5 (absolute convergence of the rearrangement). Apply Steps 1–4 verbatim to the nonnegative series $\sum |a_n|$ in place of $\sum a_n$ (a nonnegative series trivially "converges absolutely" to itself, since $||a_n|| = |a_n|$). This shows $\sum_{n=1}^\infty |a_{\sigma(n)}| = \sum_{n=1}^\infty |a_n| < \infty$, so the rearranged series converges absolutely. $\blacksquare$

Technique: proof by contradiction, splitting a series into signed parts. Suppose, for contradiction, that $\sum p_n$ converges (to a finite value $P$). Since $q_n = p_n - a_n$ and both $\sum p_n$ and $\sum a_n$ converge (the latter by hypothesis, the former by our supposition), the series $\sum q_n = \sum(p_n - a_n)$ converges as well, to $Q := P - \sum a_n$, by linearity of convergent series (a finite sum of two convergent series converges to the sum of the limits — this is immediate from the definition of a limit of partial sums and does not use anything about rearrangement). But then

again by linearity of convergent series, contradicting the hypothesis that $\sum a_n$ is only conditionally convergent (i.e. that $\sum |a_n| = \infty$). Hence $\sum p_n$ diverges; since $p_n \ge 0$, its partial sums are nondecreasing, so divergence forces $\sum p_n = +\infty$. By an identical argument with the roles of $p_n$ and $q_n$ swapped, $\sum q_n = +\infty$. $\blacksquare$

Technique: explicit greedy construction, using the Lemma above and the fact that $a_n \to 0$. First note a preliminary fact we will need: since $\sum a_n$ converges, $a_n \to 0$ as $n \to \infty$ (the terms of any convergent series tend to $0$, since $a_n = S_n - S_{n-1} \to S - S = 0$ where $S_n$ are partial sums converging to $S$). Consequently $p_n \to 0$ and $q_n \to 0$ as well, since $0 \le p_n \le |a_n|$ and $0 \le q_n \le |a_n|$. Let $(p_n)$ be the subsequence of strictly positive terms of $(a_n)$, listed in their original relative order, and let $(q_n)$ be the subsequence of the absolute values of the strictly negative terms of $(a_n)$, also listed in original relative order. (Terms equal to $0$ may be placed anywhere at the end without affecting any partial sum; discard them from this bookkeeping and reinsert them into $\sigma$ arbitrarily at the very end of the construction — this changes no partial sums past their insertion point, and formally does not disturb convergence, since adding zero terms to a partial sum leaves it unchanged.) By the Lemma applied to $\sum a_n$,

and since these are subsequences of a null sequence $a_n \to 0$, we also have $p_k \to 0$ and $q_k \to 0$.

Case 1: $L$ finite. We construct the rearrangement in blocks, alternating between runs of positive terms and runs of negative terms, using the following greedy rule. Block $1$: Since $\sum_k p_k = +\infty$ and each $p_k > 0$, there is a least integer $m_1$ such that

(Such an $m_1$ exists because the partial sums of $\sum p_k$ are nondecreasing and unbounded, hence eventually exceed $L$; "least" makes $m_1$ well defined by the well-ordering principle.) Append $p_1, \ldots, p_{m_1}$, in order, to the rearranged sequence. Let $R_1$ denote the resulting partial sum, so $R_1 > L$, and — crucially — since $m_1$ is least, we had $p_1 + \cdots + p_{m_1 - 1} \le L$, so

Block $2$: Since $\sum_k q_k = +\infty$, there is a least integer $n_1$ such that

Append $-q_1, \ldots, -q_{n_1}$, in order, to the rearranged sequence. Let $R_2$ be the resulting partial sum, so $R_2 < L$, and by leastness of $n_1$, exactly as above,

Continue inductively: having built $R_{2j-1} > L$ using the next unused block of $p$'s (least length to first exceed $L$) and $R_{2j} < L$ using the next unused block of $q$'s (least length to first go below $L$), the construction continues forever. This is possible at every stage because $\sum p_k = \infty$ and $\sum q_k = \infty$ remain true for every tail of these sequences — removing any finite number of initial terms from a series that diverges to $+\infty$ leaves a series that still diverges to $+\infty$, since only finitely much has been removed from an unbounded nondecreasing sequence of partial sums. Every term $p_k$ and every $q_k$ gets used exactly once, in exactly one block, and every block is finite, so this process defines a genuine bijection $\sigma: \mathbb{N} \to \mathbb{N}$ from the position in the newly built list to the original index — every original term appears exactly once in the new order. (Zero terms of the original series, set aside earlier, are appended in a final infinite "tail" arbitrarily reinserted at density zero without effect on limits — formally, one interleaves them so that the $N$-th partial sum of the true full rearrangement differs from the partial sum of the nonzero-only construction by a sum of finitely many zeros; we suppress this bookkeeping detail below since it changes no value.)

Convergence to $L$. We must show the partial sums of the full rearranged series converge to $L$. Consider any partial sum $T_m$ of the rearranged series where $m$ lies within block $2j-1$ (a positive-block) or block $2j$ (a negative-block). Within a positive block, partial sums are increasing from $R_{2j-2}$ up to $R_{2j-1}$ (or from $L$'s neighborhood up to $R_1$ if $j=1$); within a negative block they are decreasing from $R_{2j-1}$ down to $R_{2j}$. Hence every partial sum $T_m$ lies between the two most recent "turning" values, i.e.

More precisely and rigorously: for $m$ in the interior of block $2j-1$ (a run of positive terms), the partial sum $T_m$ satisfies $R_{2j-2} < T_m \le R_{2j-1}$ (increasing partial sums within an all-positive block, starting above or at $L$'s undershoot $R_{2j-2} < L$ and rising to overshoot $R_{2j-1} > L$); by (3)-type bound applied at stage $j$,

Symmetrically, for $m$ in block $2j$ (all-negative run), $R_{2j} \le T_m < R_{2j-1}$, and

In either case,

Since $m_j \to \infty$ and $n_j \to \infty$ as $j \to \infty$ (each block uses at least one new term, and there are infinitely many blocks since $\sum p_k, \sum q_k = \infty$ force the process never to terminate), and since $p_k \to 0$, $q_k \to 0$, the right side of (5) tends to $0$ as $j \to \infty$, i.e. as $m \to \infty$. Given $\varepsilon > 0$, choose $J$ so that $p_k, q_k < \varepsilon$ for all $k$ appearing in blocks $\ge 2J$ (possible since $p_k, q_k \to 0$); then for all $m$ past the start of block $2J$, $|T_m - L| < \varepsilon$ by (5). Hence $T_m \to L$, proving $\sum a_{\sigma(n)} = L$.

Case 2: $L = +\infty$. Modify the construction: at stage $j$, choose the least $m_j$ so that appending the next block of unused $p_k$'s makes the partial sum exceed $j$ (rather than $L$), then append exactly one unused $q_k$ (rather than a block chosen to undershoot). This is always possible: the positive terms alone sum to $+\infty$ so we can always find enough of them to clear any finite threshold $j$, and we use only a single negative term per round so the negative contribution is summable in a controlled way — concretely, since $q_k \to 0$, the single negative term used at stage $j$ contributes at most $q_k \to 0$, an amount too small to prevent the partial sums from eventually exceeding every threshold. Formally, the partial sum after stage $j$ exceeds $j - q_{k_j}$ where $q_{k_j} \to 0$, and since every original $q_k$ is used exactly once across all stages (there are infinitely many stages, one $q_k$ consumed per stage, and $\sum q_k = \infty$ only requires that we have infinitely many $q_k$ available, which we do), the construction is a valid bijection and the partial sums diverge to $+\infty$. The case $L = -\infty$ is symmetric, with the roles of $p_k$ and $q_k$ exchanged.

Oscillation. To make the rearranged series have no limit in $[-\infty,\infty]$, alternately target $L = 0$ and $L = 1$ (say): run the Case-1 construction toward $L=0$ for finitely many blocks until the partial sum is within $\tfrac13$ of $0$, then switch and run finitely many blocks of the Case-1 construction toward $L=1$ until the partial sum is within $\tfrac13$ of $1$, then switch back, forever. At each switch the argument of Case 1 (specifically bound (5), applied locally) guarantees the target is approached before switching again, so the partial sums infinitely often come within $\tfrac16$ of $0$ and infinitely often come within $\tfrac16$ of $1$; hence $\liminf T_m \le \tfrac16 < \tfrac56 \le \limsup T_m$, so $(T_m)$ has no limit in $[-\infty,\infty]$. Every original term is still used exactly once across the infinite alternation, by the same tail-never-runs-out argument as before, so $\sigma$ is a genuine bijection. $\blacksquare$