The Riemann rearrangement theorem

Theorem

The Riemann rearrangement theorem

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Statement

Let \( \sum_{n=1}^{\infty} a_n \) be a series of real numbers that converges conditionally: the partial sums \( s_N = \sum_{n=1}^{N} a_n \) converge to some \( s \in \mathbb{R} \), while \( \sum_{n=1}^{\infty} \lvert a_n \rvert = +\infty \). Then for every \( L \in \mathbb{R} \) there is a bijection \( \sigma \colon \mathbb{N} \to \mathbb{N} \) such that the rearranged series converges to \( L \), \[ \sum_{n=1}^{\infty} a_{\sigma(n)} = L . \] More generally, for every pair of extended reals \( -\infty \le \alpha \le \beta \le +\infty \) there is a bijection \( \sigma \) whose rearranged partial sums \( t_N = \sum_{n=1}^{N} a_{\sigma(n)} \) satisfy \( \liminf_{N \to \infty} t_N = \alpha \) and \( \limsup_{N \to \infty} t_N = \beta \); in particular there are rearrangements diverging to \( +\infty \), to \( -\infty \), and oscillating between any two prescribed bounds. The sum of a conditionally convergent series is therefore not a property of the terms alone, but of the terms together with their order.

Why it matters

Finite addition is commutative, and the notation \( \sum_n a_n \) invites the belief that infinite addition is too. This theorem destroys that belief in the strongest possible way: for a conditionally convergent series the set of achievable rearrangement sums is not merely larger than one point, it is the whole of \( \mathbb{R} \), together with \( \pm\infty \). Every manipulation that silently reorders terms — splitting a series into two, interchanging a double sum, summing an unordered family, integrating a series term by term after a change of variable — is illegitimate unless absolute convergence has been checked first. In practice this theorem is the reason analysis courses insist on the word “absolutely”.

It also gives the exact dividing line. Combined with its converse (Dirichlet: an absolutely convergent real series has the same sum under every rearrangement), it says that for real series unconditional convergence and absolute convergence are the same thing. That equivalence is what fails in infinite dimensions (Dvoretzky–Rogers) and is what the Lévy–Steinitz theorem replaces in \( \mathbb{R}^d \), where the set of rearrangement sums is an affine subspace rather than a point or a line-free free-for-all. Riemann's construction is thus the first theorem in the subject whose statement is about the order type of the index set rather than about sizes of terms.

Hypotheses

Hypothesis (convergence): \( \sum a_n \) converges in \( \mathbb{R} \).Convergence is used twice: it forces \( a_n \to 0 \) (the \( n \)-th term test), which is what makes the overshoot at each turning point of the construction shrink to nothing, and it forces the positive and negative parts to diverge together. Without it the conclusion can fail outright: for \( a_n = 1 \) every rearrangement has \( t_N = N \to +\infty \), so no finite \( L \) is attainable.

Hypothesis (non-absolute convergence): \( \sum \lvert a_n \rvert = +\infty \).This is the engine. It guarantees an inexhaustible supply of positive mass and of negative mass, so the running total can be driven above or below any level as often as we like. If instead \( \sum \lvert a_n \rvert \lt \infty \), Dirichlet's rearrangement theorem applies and every rearrangement converges to the same sum: \( \sum (-1)^{n+1} n^{-2} = \pi^2/12 \) whatever order the terms are taken in.

Hypothesis (real scalars): the \( a_n \) lie in \( \mathbb{R} \), with its order.The construction compares the running total with \( L \) and switches sign accordingly; that comparison needs a total order compatible with addition. In \( \mathbb{R}^2 \) the theorem is false as stated: for \( a_n = \bigl( (-1)^{n+1}/n,\, 0 \bigr) \) every rearrangement sum lies on the line \( \{ (x,0) : x \in \mathbb{R} \} \), never at \( (0,1) \). The correct \( \mathbb{R}^d \) statement is the Lévy–Steinitz theorem.

Hypothesis (rearrangement means a bijection of \( \mathbb{N} \), nothing weaker and nothing stronger).Each term must be used exactly once, and the new series is summed in its own order with no brackets. Bracketing a convergent series never changes its sum, so the theorem is not about grouping; and if \( \sigma \) is restricted so that no initial segment is scattered into more than a fixed number of pieces — precisely, if there is \( K \) such that \( \sigma(\{1,\dots,N\}) \) is a union of at most \( K \) blocks of consecutive integers for every \( N \) — then \( \sum a_{\sigma(n)} = \sum a_n \) for every convergent series (Agnew's characterisation of sum-preserving permutations). The permutations Riemann builds necessarily send some initial segments to unions of unboundedly many blocks.

Proof

Write \( x^{+} = \max(x,0) \) and \( x^{-} = \max(-x,0) \), so that \( x = x^{+} - x^{-} \) and \( \lvert x \rvert = x^{+} + x^{-} \), both parts being non-negative. We first show that the positive and the negative parts of the series both carry infinite mass (Steps 1–4), then run a greedy construction that steers the running total at \( L \) (Steps 5–11), and finally record the modifications for \( \pm\infty \) and for prescribed oscillation (Step 12).

\[ p_n := a_n^{+} = \frac{\lvert a_n \rvert + a_n}{2}, \qquad q_n := a_n^{-} = \frac{\lvert a_n \rvert - a_n}{2}, \qquad p_n, q_n \ge 0 . \]

Definition, together with the two identities \( a_n = p_n - q_n \) and \( \lvert a_n \rvert = p_n + q_n \), which are immediate from the two displayed formulas. A

Both \( \sum_n p_n \) and \( \sum_n q_n \) diverge; being series of non-negative terms, each diverges to \( +\infty \).

Suppose \( \sum p_n \lt \infty \). Since \( q_n = p_n - a_n \), the partial sums \( \sum_{n \le N} q_n = \sum_{n \le N} p_n - s_N \) converge (algebra of limits, using convergence of \( \sum a_n \)), so \( \sum q_n \lt \infty \) as well, whence \( \sum \lvert a_n \rvert = \sum (p_n + q_n) \lt \infty \) — contradicting the hypothesis of non-absolute convergence. The same argument with \( p_n = q_n + a_n \) rules out \( \sum q_n \lt \infty \). A series of non-negative terms has monotone partial sums, so divergence means divergence to \( +\infty \). B

Let \( \pi(1) \lt \pi(2) \lt \cdots \) enumerate \( \{ n : a_n \gt 0 \} \) and \( \nu(1) \lt \nu(2) \lt \cdots \) enumerate \( \{ n : a_n \lt 0 \} \); put \[ P_k := a_{\pi(k)} \gt 0, \qquad Q_k := -a_{\nu(k)} \gt 0 . \] Both index sets are infinite, and \[ \sum_{k=1}^{\infty} P_k = \sum_{n=1}^{\infty} p_n = +\infty, \qquad \sum_{k=1}^{\infty} Q_k = \sum_{n=1}^{\infty} q_n = +\infty . \]

If only finitely many \( a_n \) were positive then \( \sum p_n \) would be a finite sum, contradicting Step 2; likewise for the negatives. The two displayed identities hold because \( \sum_n p_n \) and \( \sum_k P_k \) differ only by the terms \( p_n = 0 \) (and \( \sum_n q_n \), \( \sum_k Q_k \) likewise), and deleting zero terms from a series of non-negative terms changes neither its partial sums up to reindexing nor its value in \( [0,+\infty] \). A

\[ a_n \to 0 \quad \Longrightarrow \quad P_k \to 0 \ \text{ and } \ Q_k \to 0 \quad (k \to \infty). \]

A convergent series has null terms: \( a_N = s_N - s_{N-1} \to s - s = 0 \). Since \( \pi(k) \ge k \to \infty \), the numbers \( P_k = \lvert a_{\pi(k)} \rvert \) form a subsequence of the null sequence \( ( \lvert a_n \rvert ) \), hence are null; same for \( Q_k \). This is the fact that makes the construction below converge rather than merely oscillate. B

Fix \( L \in \mathbb{R} \). Define indices recursively. Let \( m_1 \) be the least \( m \ge 1 \) with \[ \sum_{k=1}^{m} P_k \gt L , \] and, having chosen \( m_1 \lt \cdots \lt m_j \) and \( k_1 \lt \cdots \lt k_{j-1} \), let \( k_j \) be the least \( k \gt k_{j-1} \) with \[ \sum_{i=1}^{m_j} P_i - \sum_{i=1}^{k} Q_i \lt L , \] and then \( m_{j+1} \) the least \( m \gt m_j \) with \( \sum_{i=1}^{m} P_i - \sum_{i=1}^{k_j} Q_i \gt L \).

The greedy rule in words: pile on positive terms, in their original relative order, until the running total first exceeds \( L \); then pile on negative terms until it first falls below \( L \); repeat forever. B

Every index in Step 5 exists: the defining sets are non-empty.

Because \( \sum_k P_k = +\infty \) (Step 3), the quantity \( \sum_{i=1}^{m} P_i - \sum_{i=1}^{k_j} Q_i \) tends to \( +\infty \) as \( m \to \infty \) with \( k_j \) fixed, so it exceeds \( L \) for some \( m \gt m_j \); because \( \sum_k Q_k = +\infty \), the quantity \( \sum_{i=1}^{m_j} P_i - \sum_{i=1}^{k} Q_i \) tends to \( -\infty \) as \( k \to \infty \), so it falls below \( L \) for some \( k \gt k_{j-1} \). The recursion therefore never stalls, and produces strictly increasing sequences \( (m_j) \) and \( (k_j) \) with \( m_j \to \infty \), \( k_j \to \infty \). B

Define \( \sigma \) by listing, in this order, \[ \underbrace{P_1,\dots,P_{m_1}}_{\text{block } 1^{+}},\ \underbrace{-Q_1,\dots,-Q_{k_1}}_{\text{block } 1^{-}},\ \underbrace{P_{m_1+1},\dots,P_{m_2}}_{\text{block } 2^{+}},\ \underbrace{-Q_{k_1+1},\dots,-Q_{k_2}}_{\text{block } 2^{-}},\ \dots \] and inserting the terms \( a_n = 0 \) (if any) one per block, in increasing order of \( n \), at the end of each block; if there are only finitely many zero terms, place them all at the very front.

Every original index appears exactly once: the positive indices \( \pi(1),\pi(2),\dots \) are used in order and \( m_j \to \infty \) exhausts them; the negative indices likewise since \( k_j \to \infty \); the zero indices are exhausted by the block-by-block insertion. Hence \( \sigma \) is a bijection \( \mathbb{N} \to \mathbb{N} \). Zero terms leave every partial sum unchanged, so they are ignored from here on. B

Let \( u_j \) be the rearranged partial sum at the end of block \( j^{+} \) and \( v_j \) the one at the end of block \( j^{-} \). Then \[ 0 \lt u_j - L \le P_{m_j}, \qquad 0 \lt L - v_j \le Q_{k_j} . \]

Take \( j \ge 2 \). The block \( j^{+} \) starts from \( v_{j-1} \lt L \) and adds positive terms; by minimality of \( m_j \) the partial sum one term earlier, \( u_j - P_{m_j} \), does not exceed \( L \), so \( u_j - P_{m_j} \le L \lt u_j \), which is the first inequality. By minimality of \( k_j \) the sum one term before \( v_j \) is \( \ge L \), i.e. \( v_j + Q_{k_j} \ge L \gt v_j \), which is the second. (The very first block is the only exception: if \( L \lt 0 \) then \( m_1 = 1 \) and \( u_1 - L = P_1 - L \) can exceed \( P_1 \). Nothing is lost, because the estimate is used in Step 10 only for large \( j \).) C

Within block \( j^{+} \) the partial sums increase from \( v_{j-1} \) to \( u_j \); within block \( j^{-} \) they decrease from \( u_j \) to \( v_j \). Hence for every \( N \) whose position lies in block \( j^{+} \) or \( j^{-} \) (with \( j \ge 2 \)), \[ \min\bigl( v_{j-1},\, v_j \bigr) \le t_N \le u_j , \qquad \text{so} \qquad \lvert t_N - L \rvert \le \max\bigl( P_{m_j},\, Q_{k_{j-1}},\, Q_{k_j} \bigr) . \]

Adding positive numbers increases a partial sum and subtracting positive numbers decreases it, so within a block the sequence \( (t_N) \) is monotone and therefore trapped between its endpoint values: block \( j^{+} \) runs from \( v_{j-1} \) up to \( u_j \), block \( j^{-} \) from \( u_j \) down to \( v_j \). Since \( v_{j-1} \lt L \lt u_j \) and \( v_j \lt L \lt u_j \), the target \( L \) lies inside both intervals, so \( \lvert t_N - L \rvert \le \max\bigl( u_j - L,\ L - v_{j-1},\ L - v_j \bigr) \), and Step 8 bounds those three quantities by \( P_{m_j} \), \( Q_{k_{j-1}} \) and \( Q_{k_j} \) respectively. B

\[ \lim_{N \to \infty} t_N = L , \qquad \text{i.e.} \qquad \sum_{n=1}^{\infty} a_{\sigma(n)} = L . \]

Let \( \varepsilon \gt 0 \). By Step 4 there is \( J \ge 2 \) with \( P_{m_j} \lt \varepsilon \) and \( Q_{k_j} \lt \varepsilon \) for all \( j \ge J \) (using \( m_j, k_j \to \infty \) from Step 6). Every position \( N \) beyond the end of block \( J^{-} \) lies in some block \( j^{\pm} \) with \( j \ge J+1 \), so both \( j \) and \( j-1 \) are \( \ge J \) and Step 9 gives \( \lvert t_N - L \rvert \le \max(P_{m_j}, Q_{k_{j-1}}, Q_{k_j}) \lt \varepsilon \). That is the \( \varepsilon \)–\( N \) definition of \( t_N \to L \). B

The permutation depends on \( L \): distinct targets \( L \ne L' \) give rearrangements with \( \sum a_{\sigma(n)} = L \) and \( \sum a_{\sigma'(n)} = L' \), so the map “order \( \mapsto \) sum” is onto \( \mathbb{R} \).

Steps 5–10 were carried out for an arbitrary fixed \( L \in \mathbb{R} \); nothing in them used any property of \( L \). Surjectivity onto \( \mathbb{R} \) is the assertion of the theorem's first sentence. A

Prescribed \( \liminf \) and \( \limsup \). Given \( -\infty \le \alpha \le \beta \le +\infty \), choose real sequences \( \alpha_j \to \alpha \) and \( \beta_j \to \beta \) with \( \alpha_j \lt \beta_j \) for all \( j \), and run the same greedy scheme with a moving target: positive terms until the total first exceeds \( \beta_j \), then negative terms until it first falls below \( \alpha_j \), then on to \( \beta_{j+1} \). The resulting \( (t_N) \) satisfies \[ \limsup_{N \to \infty} t_N = \beta, \qquad \liminf_{N \to \infty} t_N = \alpha . \] Taking \( \alpha = \beta = +\infty \) (say \( \beta_j = j+1 \), \( \alpha_j = j \)) yields a rearrangement diverging to \( +\infty \); \( \alpha = -1 \), \( \beta = 1 \) yields one oscillating with those exact bounds.

Each stage terminates for the reason given in Step 6, and the overshoot at the \( j \)-th top turning point is at most the last positive term used there, which tends to \( 0 \) by Step 4; hence the top turning values converge to \( \beta \) and the bottom ones to \( \alpha \). Between turning points the partial sums are monotone, so they contribute no subsequential limits outside \( [\alpha,\beta] \) beyond those already accounted for; every subsequential limit therefore lies in \( [\alpha, \beta] \) and both endpoints are attained along the turning points. C

Result

\[ \sum_n a_n \ \text{converges},\ \ \sum_n \lvert a_n \rvert = \infty \quad \Longrightarrow \quad \forall\, L \in \mathbb{R} \cup \{\pm\infty\}\ \ \exists\, \sigma \ \text{bijective}: \ \sum_{n=1}^{\infty} a_{\sigma(n)} = L . \]

Reading. A conditionally convergent series has an unlimited reserve of positive mass and an unlimited reserve of negative mass, drawn from terms that are individually shrinking to zero. Spend the reserves greedily — positives until you are above your target, negatives until you are below — and the running total is squeezed onto that target, because each overshoot is at most the size of the last term used, and those sizes vanish. The sum of the series is whatever you decide it should be.

Scope. Real scalars, a genuine bijection of the index set, and non-absolute convergence. For absolutely convergent series the conclusion reverses completely (Dirichlet: the sum is rearrangement-invariant), and for \( \mathbb{R}^d \)-valued series it is replaced by the Lévy–Steinitz theorem, which says the set of rearrangement sums is a translate of a linear subspace. The proof is constructive but the permutation is defined by an unbounded search, not by a closed formula, and it depends on \( L \).

Corollaries & converses

Dichotomy for the sum set. For a convergent real series let \( S = \bigl\{ \sum_n a_{\sigma(n)} : \sigma \ \text{a bijection of } \mathbb{N},\ \text{the series converging} \bigr\} \). Then either \( \sum \lvert a_n \rvert \lt \infty \) and \( S \) is a single point, or \( S = \mathbb{R} \). Nothing in between is possible.
Converse (Dirichlet). If \( \sum \lvert a_n \rvert \lt \infty \) then every rearrangement converges to the same sum. Hence for real series unconditional convergence \( \iff \) absolute convergence, and Riemann's theorem is exactly the non-trivial half.
Divergent rearrangements. Taking \( \alpha = \beta = +\infty \) in Step 12 gives \( \sigma \) with \( t_N \to +\infty \); similarly \( -\infty \). So a convergent series can be rearranged into a divergent one precisely when it is not absolutely convergent.
Block rearrangement of the alternating harmonic series. Taking \( p \) positive terms then \( q \) negative terms, repeatedly, gives \[ 1 + \tfrac13 + \cdots - \tfrac12 - \cdots \ \longrightarrow \ \ln 2 + \tfrac12 \ln\!\left( \frac{p}{q} \right), \] so \( (p,q) = (2,1) \) gives \( \tfrac32 \ln 2 \approx 1.039721 \) and \( (1,2) \) gives \( \tfrac12 \ln 2 \approx 0.346574 \).
No unordered sum. A family \( (a_n)_{n \in \mathbb{N}} \) of reals is summable in the net sense (finite partial sums over finite subsets converging along the inclusion-ordered net) if and only if \( \sum \lvert a_n \rvert \lt \infty \). Conditionally convergent series have no order-free value at all.
Lévy–Steinitz (\( \mathbb{R}^d \)). For a convergent series in \( \mathbb{R}^d \), the set of sums of convergent rearrangements is \( s + \Gamma^{\perp} \) for a linear subspace \( \Gamma^{\perp} \) — a point when the series converges absolutely, all of \( \mathbb{R}^d \) at the other extreme, and any intermediate affine subspace in between. Riemann's theorem is the case \( d = 1 \).
Dvoretzky–Rogers. In every infinite-dimensional Banach space there exists a series that converges unconditionally but not absolutely, so the equivalence “unconditional \( = \) absolute” is a strictly finite-dimensional phenomenon.

Fails without

Absolute convergence instead of conditional: for \( a_n = (-1)^{n+1}/n^2 \) we have \( \sum \lvert a_n \rvert = \pi^2/6 \lt \infty \), and every rearrangement converges to \( \pi^2/12 \approx 0.822467 \). The greedy construction cannot be run indefinitely: once the positive reserve \( \sum_k P_k = \pi^2/8 \approx 1.233700 \) is exhausted, the running total can never be pushed above \( L = 2 \) again.
Vector-valued terms: in \( \mathbb{R}^2 \) put \( a_n = \bigl( (-1)^{n+1}/n,\, 0 \bigr) \). The series converges (to \( (\ln 2, 0) \)) and not absolutely, since \( \sum \lVert a_n \rVert = \sum 1/n = \infty \), yet every rearrangement sum has second coordinate \( 0 \). The attainable set is the line \( \mathbb{R} \times \{0\} \), not \( \mathbb{R}^2 \): the conclusion “any prescribed value” is false, and Lévy–Steinitz is the correct replacement.
Bounded-block permutations: if \( \sigma \) is such that each \( \sigma(\{1,\dots,N\}) \) is a union of at most \( K \) blocks of consecutive integers (\( K \) independent of \( N \)) then \( \sum a_{\sigma(n)} = \sum a_n \) for every convergent series. Swapping each pair \( (2k-1,2k) \) of the alternating harmonic series gives \( K = 2 \) (an initial segment of odd length is \( \{1,\dots,N-1\} \cup \{N+1\} \)) and the sum stays stubbornly \( \ln 2 \). Unbounded reordering is essential, not incidental.
Bracketing instead of permuting: inserting brackets into a convergent series never changes its value, since the bracketed partial sums form a subsequence of \( (s_N) \). So \( (1 - \tfrac12) + (\tfrac13 - \tfrac14) + \cdots = \ln 2 \) still. (Bracketing can only manufacture convergence out of divergence, as in \( (1-1) + (1-1) + \cdots = 0 \).)
Terms of one sign: if \( a_n \ge 0 \) for all \( n \), then \( \sum_n a_{\sigma(n)} = \sum_n a_n \) in \( [0,+\infty] \) for every bijection \( \sigma \), because both sides equal \( \sup \bigl\{ \sum_{n \in F} a_n : F \subseteq \mathbb{N} \ \text{finite} \bigr\} \). With no negative reserve there is nothing to steer with.

Common errors

“The terms are the same, so the sum is the same.” Commutativity is a statement about finite sums; an infinite sum is a limit of a sequence of partial sums, and permuting the terms produces a different sequence. Riemann's theorem is precisely the failure of the induction from finite to infinite.
Confusing rearrangement with grouping. Bracketing a convergent series preserves its sum (bracketed sums are a subsequence of the partial sums). The classic “\( 1 - 1 + 1 - 1 + \cdots = 0 \) or \( 1 \)” paradox is about a divergent series and has nothing to do with this theorem.
Expecting a formula for \( \sigma \). The construction is a greedy search whose block lengths depend on the target \( L \) and on the arithmetic of the terms; there is no closed form in general. Only for structured examples (the \( p \):\( q \) blocks of the alternating harmonic series) is the permutation explicit.
Believing that any specific rearrangement changes the sum. The theorem asserts existence, not typicality. Most permutations you write down — certainly all bounded-block ones — leave the sum alone; \( \sum (-1)^{n+1}/n \) with each adjacent pair swapped is still \( \ln 2 \).
Applying it to a series that is not conditionally convergent. “\( \sum 1/n \) can be rearranged to converge to \( 7 \)” is false: the theorem needs the original series to converge. A divergent series of non-negative terms diverges under every rearrangement.
Numerical “verification” that a rearrangement has the original sum. Rearranged conditionally convergent series converge extremely slowly — the error decays like \( 1/n \), not geometrically. The \( 2\):\(1 \) rearrangement of the alternating harmonic series has \( t_{3n} = 1.015189 \) at \( n = 10 \) and \( 1.037225 \) at \( n = 100 \), and even \( t_{3000} = 1.039471 \) has not yet settled the fourth decimal of \( \tfrac32\ln 2 = 1.039721 \). A short numerical run therefore proves nothing either way: it cannot confirm a limit, and a drifting partial sum is not evidence that the sum has been preserved.
Assuming absolute convergence lets you do anything. It licenses rearrangement and Cauchy products of real series, but not interchange of limits in general; that needs its own hypotheses (dominated convergence, uniform convergence).

Discussion

Riemann proved this in his 1854 Habilitationsschrift on the representability of a function by a trigonometric series, published posthumously in 1867. The theorem sits there as a lemma-sized remark, but it is the moment when “sum of a series” stopped being a property of a collection of numbers and became a property of a collection plus an order. Dirichlet had already noticed in the 1830s, while working on Fourier series and on primes in arithmetic progressions, that reordering a non-absolutely convergent series could change its value; Cauchy's earlier treatments of series manipulation quietly presuppose absolute convergence throughout. Riemann's contribution was to show that the pathology is total: not some other value, but every value.

The mechanism is worth isolating, because it recurs. Conditional convergence means the series is convergent for a reason of cancellation, not of smallness: the positive part and the negative part are each infinite, and the finite answer is the difference of two infinities, stabilised only by the particular order in which they are interleaved. Change the interleaving and you change the answer; that is all the theorem says. The same “\( \infty - \infty \)” structure is why the Lebesgue integral is defined only for functions with \( \int \lvert f \rvert \lt \infty \), why Fubini's theorem carries an integrability hypothesis (Tonelli's version needs no such hypothesis precisely because the integrand is non-negative), and why conditionally convergent improper integrals such as \( \int_0^\infty \frac{\sin x}{x}\, dx = \frac{\pi}{2} \) do not survive arbitrary substitutions of the domain.

The result also fixes the meaning of a notation. In the theory of summable families one defines \( \sum_{n \in I} a_n \) for an arbitrary index set \( I \) as a limit over finite subsets, with no order at all; the family is summable exactly when \( \sum \lvert a_n \rvert \lt \infty \). So “\( \sum_{n=1}^{\infty} (-1)^{n+1}/n = \ln 2 \)” is not a statement about the set of numbers \( \{ \pm 1/n \} \) at all — it is a statement about the sequence in which they are presented. Once a student has internalised that, a whole class of false manipulations (splitting a conditionally convergent series into its positive and negative parts and summing each; reindexing a double series without checking absolute summability) becomes visibly illegal.

Two refinements sharpen the picture. In \( \mathbb{R}^d \), Lévy (1905) and Steinitz (1913) showed that the set of rearrangement sums of a convergent series is always an affine subspace \( s + \Gamma^{\perp} \), where \( \Gamma \) is the set of linear functionals \( f \) with \( \sum_n \lvert f(a_n) \rvert \lt \infty \); Riemann's theorem is the case \( d = 1 \), where \( \Gamma \) is either \( \{0\} \) (giving \( \Gamma^{\perp} = \mathbb{R} \)) or all of \( \mathbb{R}^{*} \) (giving a single point). In infinite dimensions the dichotomy breaks completely: by Dvoretzky–Rogers (1950) every infinite-dimensional Banach space carries an unconditionally but not absolutely convergent series — in \( \ell^2 \), for instance, \( a_n = e_n / n \) converges unconditionally while \( \sum \lVert a_n \rVert = \sum 1/n = \infty \). Finally, on the permutation side, Agnew characterised the bijections that preserve the sum of every convergent real series: they are exactly those for which \( \sigma(\{1,\dots,N\}) \) is a union of at most \( K \) blocks of consecutive integers, with \( K \) independent of \( N \). Riemann's permutations must therefore scatter initial segments into unboundedly many blocks, and the construction shows exactly how. In Example 2 below (target \( L = 0 \)) each stage consumes one odd index and four even ones, so after \( j \) stages the image of the initial segment of length \( 5j \) is \( \{1,\dots,2j\} \cup \{2j+2, 2j+4, \dots, 8j\} \) — one interval followed by \( 3j \) isolated points, that is \( 3j+1 \) blocks, growing without bound exactly as Agnew's criterion demands.

Common misconceptions. The theorem does not say that a conditionally convergent series has no sum — it has exactly one, in the order given. It does not say that rearranging usually changes the sum; it says a suitable rearrangement can achieve any target. And it does not extend to absolutely convergent series in any weakened form: there the sum really is a function of the multiset of terms alone.

Worked examples

Example 1. The alternating harmonic series \( \sum_{n=1}^{\infty} (-1)^{n+1}/n \) converges to \( \ln 2 = 0.693147\ldots \). Take its terms in the order “two positives, one negative”, \[ 1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \tfrac19 + \tfrac1{11} - \tfrac16 + \cdots , \] and evaluate the resulting sum exactly.

The hypotheses hold: \( \sum (-1)^{n+1}/n \) converges (alternating series test) and \( \sum 1/n = \infty \), so the series is conditionally convergent and Riemann's theorem applies.

Both facts are standard: monotone null terms give convergence; the harmonic series diverges. So a change of value under rearrangement is possible — we now compute the actual value for this particular order. A

\( \sigma \) is a bijection: the positive terms \( 1, \tfrac13, \tfrac15, \dots \) appear in their original relative order, two at a time, and the negative terms \( -\tfrac12, -\tfrac14, -\tfrac16, \dots \) one at a time, so each original index is used exactly once.

Position \( 3j-2, 3j-1 \) carries \( \tfrac{1}{4j-3}, \tfrac{1}{4j-1} \) and position \( 3j \) carries \( -\tfrac{1}{2j} \); as \( j \) runs over \( \mathbb{N} \) these exhaust the odd and the even denominators respectively. A

\[ t_{3n} = \sum_{j=1}^{n} \left( \frac{1}{4j-3} + \frac{1}{4j-1} - \frac{1}{2j} \right) = \sum_{j=1}^{2n} \frac{1}{2j-1} \; - \; \frac{1}{2} H_n , \qquad H_m := \sum_{k=1}^{m} \frac1k . \]

Regrouping a finite sum is legitimate. The first \( 2n \) odd reciprocals are exactly the positive terms used; the \( n \) negative terms are \( \tfrac12 (1 + \tfrac12 + \cdots + \tfrac1n) = \tfrac12 H_n \). A

\[ \sum_{j=1}^{2n} \frac{1}{2j-1} = H_{4n} - \frac{1}{2} H_{2n} \qquad \Longrightarrow \qquad t_{3n} = H_{4n} - \tfrac12 H_{2n} - \tfrac12 H_n . \]

Split \( H_{4n} \) into odd and even denominators: \( H_{4n} = \sum_{j=1}^{2n} \frac{1}{2j-1} + \sum_{j=1}^{2n} \frac{1}{2j} \) and the even part is \( \tfrac12 H_{2n} \). Check at \( n=1 \): \( H_4 - \tfrac12 H_2 = \tfrac{25}{12} - \tfrac34 = \tfrac43 = 1 + \tfrac13 \). B

Insert \( H_m = \ln m + \gamma + \varepsilon_m \) with \( \varepsilon_m \to 0 \) and \( \gamma = 0.5772156649\ldots \) (Euler–Mascheroni): \[ t_{3n} = \ln(4n) - \tfrac12 \ln(2n) - \tfrac12 \ln n + \gamma\left( 1 - \tfrac12 - \tfrac12 \right) + o(1) . \]

Standard asymptotics of the harmonic numbers; the three copies of \( \gamma \) cancel exactly because the coefficients \( 1, -\tfrac12, -\tfrac12 \) sum to \( 0 \) — which is the same bookkeeping that makes the \( \ln n \) terms cancel. B

\[ \ln(4n) - \tfrac12\ln(2n) - \tfrac12\ln n = \ln 4 + \ln n - \tfrac12 \ln 2 - \tfrac12 \ln n - \tfrac12 \ln n = 2\ln 2 - \tfrac12 \ln 2 = \tfrac32 \ln 2 . \]

Logarithm laws; every \( \ln n \) cancels, leaving a constant. Hence \( t_{3n} \to \tfrac32 \ln 2 = 1.039721\ldots \). B

\( t_{3n-1} = t_{3n} + \tfrac{1}{2n} \) and \( t_{3n-2} = t_{3n} + \tfrac1{2n} - \tfrac{1}{4n-1} \), both differing from \( t_{3n} \) by \( O(1/n) \to 0 \), so the full sequence \( (t_N) \) converges to the same limit.

The three residue classes of \( N \) modulo \( 3 \) give subsequences with a common limit, so \( t_N \to \tfrac32 \ln 2 \) (a sequence converges iff finitely many subsequences covering all indices converge to one value). Numerically: \( t_3 = 0.833333 \), \( t_6 = 0.926190 \), \( t_{30} = 1.015189 \), \( t_{300} = 1.037225 \), \( t_{3000} = 1.039471 \). B

\[ 1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \cdots = \tfrac32 \ln 2 \approx 1.039721 \quad \ne \quad \ln 2 \approx 0.693147 . \]

Reading. The same terms, each used exactly once, in a different order, sum to exactly \( 1.5 \) times the original value. The extra \( \tfrac12 \ln 2 \) is the accumulated advantage of spending positive terms twice as fast as negative ones.

Scope. Exact, and the method generalises: \( p \) positives to \( q \) negatives gives \( \ln 2 + \tfrac12 \ln(p/q) \). It relies on the harmonic asymptotics, not on Riemann's construction — the theorem guarantees that some order gives any target; here we evaluate one specific order.

Example 2. Run Riemann's construction on the same series with target \( L = 0 \): exhibit the first three stages explicitly, verify the overshoot bound of Step 8 numerically, and state the resulting permutation.

Positive reserve \( P_k = \dfrac{1}{2k-1} \) (values \( 1, \tfrac13, \tfrac15, \dots \)); negative reserve \( Q_k = \dfrac{1}{2k} \) (values \( \tfrac12, \tfrac14, \tfrac16, \dots \)). Both sum to \( +\infty \) and both are null.

\( \sum_k \frac{1}{2k-1} \ge \sum_k \frac{1}{2k} = \tfrac12 \sum 1/k = \infty \), and \( \sum_k \frac{1}{2k} = \infty \) likewise; \( P_k, Q_k \to 0 \). Steps 3–4 of the proof are verified concretely. A

Stage 1, positives: the least \( m \) with \( \sum_{k \le m} P_k \gt 0 \) is \( m_1 = 1 \). Running total \( u_1 = 1 \).

One term already overshoots the target \( L = 0 \). Overshoot \( u_1 - L = 1 = P_{m_1} \), consistent with Step 8. A

Stage 1, negatives: subtract \( \tfrac12, \tfrac14, \tfrac16, \tfrac18 \). \[ 1 - \tfrac12 = 0.500000, \quad -\tfrac14 \to 0.250000, \quad -\tfrac16 \to 0.083333, \quad -\tfrac18 \to -\tfrac{1}{24} = -0.041667 \lt 0 . \] So \( k_1 = 4 \) and \( v_1 = -1/24 \).

Greedy rule: stop at the first term that takes the total below \( L = 0 \). Exactly: \( 1 - \tfrac12(1 + \tfrac12 + \tfrac13 + \tfrac14) = 1 - \tfrac{25}{24} = -\tfrac{1}{24} \). Undershoot \( L - v_1 = 0.041667 \le Q_{k_1} = \tfrac18 = 0.125 \), as Step 8 requires. B

Stage 2: add \( P_2 = \tfrac13 \), giving \( u_2 = -\tfrac1{24} + \tfrac13 = \tfrac{7}{24} = 0.291667 \gt 0 \) (so \( m_2 = 2 \)); then subtract \( \tfrac1{10}, \tfrac1{12}, \tfrac1{14}, \tfrac1{16} \): \[ 0.191667,\quad 0.108333,\quad 0.036905,\quad -0.025595 \lt 0, \] so \( k_2 = 8 \), \( v_2 = -0.025595 \).

Arithmetic. Overshoot \( u_2 - L = 0.291667 \le P_{m_2} = \tfrac13 = 0.333333 \); undershoot \( 0.025595 \le Q_{k_2} = \tfrac1{16} = 0.0625 \). Both bounds of Step 8 hold with room to spare. B

Stage 3: add \( P_3 = \tfrac15 \) to get \( u_3 = 0.174405 \); subtract \( \tfrac1{18}, \tfrac1{20}, \tfrac1{22}, \tfrac1{24} \) to get \( 0.118849,\ 0.068849,\ 0.023395,\ v_3 = -0.018272 \). Hence for every position \( N \) at or beyond stage 3, \[ \lvert t_N \rvert \le \max\left( P_{m_3},\, Q_{k_3} \right) = \max\left( \tfrac15, \tfrac1{24} \right) = 0.2 . \]

Step 9 applied with \( j = 3 \): inside a block the partial sums are monotone, so they stay between \( v_3 \) and \( u_3 \), and both lie within \( \max(P_{m_3}, Q_{k_3}) \) of \( L = 0 \). The bound tightens to \( P_{m_j} \to 0 \) as the stages proceed. B

The permutation begins \[ 1,\ -\tfrac12,\ -\tfrac14,\ -\tfrac16,\ -\tfrac18,\ \tfrac13,\ -\tfrac1{10},\ -\tfrac1{12},\ -\tfrac1{14},\ -\tfrac1{16},\ \tfrac15,\ -\tfrac1{18},\ -\tfrac1{20},\ -\tfrac1{22},\ -\tfrac1{24},\ \tfrac17,\ \dots \] and \( t_N \to 0 \) by Step 10.

Each odd denominator appears once (one per stage, in order) and each even denominator appears once (four per stage in the stages above, and in every case a finite run taken in order), so \( \sigma \) is a bijection; the error bound \( \max(P_{m_j}, Q_{k_j}) \to 0 \) forces convergence to the target. A

\[ \sum_{n=1}^{\infty} a_{\sigma(n)} = 0 \quad \text{while} \quad \sum_{n=1}^{\infty} a_n = \ln 2 = 0.693147\ldots, \qquad \lvert t_N \rvert \le 0.2 \ \text{ from stage } 3 \text{ on}. \]

Reading. Four negative terms per positive term is enough to hold the running total at zero, because the negative reserve \( \tfrac12 + \tfrac14 + \cdots \) is also infinite; the cost of each course correction is the size of the last term used, and that decays like \( 1/N \).

Scope. The same recipe with \( L = 2 \) or \( L = -100 \) works verbatim, only with different block lengths — for \( L = 2 \) the first block needs the eight positive terms \( 1, \tfrac13, \dots, \tfrac1{15} \), whose sum is \( 2.021800 \). No feature of the target was used.

Problems

Let \( \sum a_n \) converge conditionally. Prove that \( \sum_n a_n^{+} = \sum_n a_n^{-} = +\infty \), that infinitely many terms are \( \gt 0 \) and infinitely many are \( \lt 0 \), and that the sequences \( (P_k) \) and \( (Q_k) \) of Step 3 are null.
Solution

Write \( p_n = a_n^{+} \), \( q_n = a_n^{-} \), so \( p_n, q_n \ge 0 \), \( a_n = p_n - q_n \) and \( \lvert a_n \rvert = p_n + q_n \). Suppose \( \sum p_n \) converged, with value \( A \). Since \( q_n = p_n - a_n \), the partial sums satisfy \( \sum_{n \le N} q_n = \sum_{n \le N} p_n - s_N \to A - s \), so \( \sum q_n \) converges too, and then \( \sum \lvert a_n \rvert = \sum p_n + \sum q_n = A + (A - s) \lt \infty \), contradicting non-absolute convergence. Symmetrically, \( p_n = q_n + a_n \) shows \( \sum q_n \lt \infty \) is impossible. Both series have non-negative terms, hence monotone partial sums, so divergence means divergence to \( +\infty \).

If only finitely many \( a_n \) were positive, say all for \( n \le N_0 \), then \( \sum_n p_n = \sum_{n \le N_0} p_n \lt \infty \), contradiction; likewise for negatives. Finally, convergence of \( \sum a_n \) gives \( a_N = s_N - s_{N-1} \to s - s = 0 \), so \( \lvert a_n \rvert \to 0 \). Since \( \pi(k) \ge k \) and \( \nu(k) \ge k \), the sequences \( P_k = \lvert a_{\pi(k)} \rvert \) and \( Q_k = \lvert a_{\nu(k)} \rvert \) are subsequences of a null sequence, hence null.
Evaluate the rearrangement of \( \sum (-1)^{n+1}/n \) that takes three positive terms then one negative term, repeatedly: \( 1 + \tfrac13 + \tfrac15 - \tfrac12 + \tfrac17 + \tfrac19 + \tfrac1{11} - \tfrac14 + \cdots \). Give the value to six decimal places and compare with a numerical partial sum.
Solution

After \( n \) blocks (that is, \( 4n \) terms) the positive terms used are the first \( 3n \) odd reciprocals and the negative terms are \( -\tfrac12, \dots, -\tfrac1{2n} \). Hence \[ t_{4n} = \sum_{j=1}^{3n} \frac{1}{2j-1} - \frac{1}{2} H_n = \left( H_{6n} - \tfrac12 H_{3n} \right) - \tfrac12 H_n , \] using \( \sum_{j=1}^{m} \frac{1}{2j-1} = H_{2m} - \tfrac12 H_m \) with \( m = 3n \). Substituting \( H_m = \ln m + \gamma + o(1) \), the \( \gamma \) coefficients \( 1 - \tfrac12 - \tfrac12 \) cancel and \[ t_{4n} \to \ln(6n) - \tfrac12 \ln(3n) - \tfrac12 \ln n = \ln 6 - \tfrac12 \ln 3 = \ln\frac{6}{\sqrt3} = \ln\bigl( 2\sqrt3 \bigr) = \ln 2 + \tfrac12 \ln 3 . \] Numerically \( \ln 2 + \tfrac12 \ln 3 = 0.693147 + 0.549306 = 1.242453 \). The intermediate partial sums \( t_{4n-1}, t_{4n-2}, t_{4n-3} \) differ from \( t_{4n} \) by at most \( \tfrac{1}{2n} + \tfrac{2}{6n-5} \to 0 \), so the whole sequence converges to the same value. As a check, summing \( 800 \) terms (\( n = 200 \)) gives \( t_{800} = 1.241204 \), approaching \( 1.242453 \) from below at the expected rate \( O(1/n) \). This matches the general formula \( \ln 2 + \tfrac12 \ln (p/q) \) with \( p = 3 \), \( q = 1 \).
Show that \( \sum_{n=1}^{\infty} (-1)^{n+1} n^{-1/2} \) is conditionally convergent, and construct explicitly a rearrangement whose partial sums diverge to \( +\infty \). Give the first two blocks numerically.
Solution

Conditional convergence. \( a_n = n^{-1/2} \) is positive, decreasing and null, so the alternating series test gives convergence; \( \sum n^{-1/2} \) is a \( p \)-series with \( p = \tfrac12 \le 1 \), hence divergent, so the convergence is not absolute.

Construction. Positives are \( P_k = (2k-1)^{-1/2} \) and negatives \( -Q_j \) with \( Q_j = (2j)^{-1/2} \). Since \( \sum_k P_k = \infty \), we may choose blocks recursively: let block \( j \) consist of the next consecutive positive terms, taken until their sum exceeds \( 1 + Q_j \), and follow it by the single negative term \( -Q_j \). Let \( u_j \) be the partial sum at the end of block \( j \) together with its negative term. Then \[ u_j \gt u_{j-1} + (1 + Q_j) - Q_j = u_{j-1} + 1, \] so \( u_j \gt j \) by induction (\( u_0 = 0 \)). Within block \( j \) the partial sums only increase from \( u_{j-1} \), and the single negative step takes them to \( u_j \gt u_{j-1} \); hence every partial sum after block \( j-1 \) is at least \( u_{j-1} \gt j-1 \), and \( t_N \to +\infty \). Every term is used exactly once because both index lists are consumed in order and the blocks are finite.

Numerics. Block 1 needs the positive sum to exceed \( 1 + Q_1 = 1 + 2^{-1/2} = 1.707107 \): \( 1 + 3^{-1/2} = 1.577350 \) is not enough, and \( 1 + 3^{-1/2} + 5^{-1/2} = 2.024564 \) is, so block 1 is \( 1, \tfrac{1}{\sqrt3}, \tfrac{1}{\sqrt5} \) followed by \( -\tfrac{1}{\sqrt2} \), leaving \( u_1 = 1.317457 \gt 1 \). Block 2 needs an increase past \( 1 + Q_2 = 1.5 \): the four terms \( 7^{-1/2}, \dots, 13^{-1/2} \) sum to only \( 1.290159 \), while \( 7^{-1/2} + 9^{-1/2} + 11^{-1/2} + 13^{-1/2} + 15^{-1/2} = 1.548358 \) suffices, so block 2 is those five terms followed by \( -\tfrac12 \), giving \( u_2 = 2.365815 \gt 2 \). The following block lengths are \( 7, 9, 10, 12, 13, 15, \dots \): they grow roughly linearly in \( j \), because \( \sum_{k \le K} (2k-1)^{-1/2} \approx \sqrt{2K} \) must increase by about \( 1 \) per block, so the number \( K_j \) of positive terms used after \( j \) blocks is of order \( j^2 \) and the \( j \)-th block contributes \( K_j - K_{j-1} = O(j) \) of them.
Prove the converse (Dirichlet): if \( \sum_n \lvert a_n \rvert \lt \infty \) then for every bijection \( \sigma \colon \mathbb{N} \to \mathbb{N} \) the series \( \sum_n a_{\sigma(n)} \) converges, with the same sum \( s \).
Solution

Let \( \varepsilon \gt 0 \). By convergence of \( \sum \lvert a_n \rvert \) (Cauchy criterion) choose \( M \) with \( \sum_{n \gt M} \lvert a_n \rvert \lt \varepsilon/2 \), and enlarge \( M \) if necessary so that \( \lvert s_M - s \rvert \lt \varepsilon/2 \). The finite set \( \{1,\dots,M\} \) has a finite preimage under \( \sigma \), so we may set \( N_0 = \max \{ \sigma^{-1}(1), \dots, \sigma^{-1}(M) \} \). For every \( N \ge N_0 \), the index set \( \sigma(\{1,\dots,N\}) \) contains \( \{1,\dots,M\} \), so \[ \lvert t_N - s_M \rvert = \Bigl\lvert \sum_{n \in \sigma(\{1,\dots,N\}) \setminus \{1,\dots,M\}} a_n \Bigr\rvert \le \sum_{n \gt M} \lvert a_n \rvert \lt \frac{\varepsilon}{2}, \] where the middle expression is a finite sum of terms with indices \( \gt M \) (the triangle inequality is applied to a finite sum, and the bound uses only that those indices are distinct and exceed \( M \)). Therefore \( \lvert t_N - s \rvert \le \lvert t_N - s_M \rvert + \lvert s_M - s \rvert \lt \varepsilon \) for all \( N \ge N_0 \), i.e. \( t_N \to s \). Note where the hypothesis is used: the tail bound must control an arbitrary finite collection of far-out indices, which requires the absolute tail \( \sum_{n \gt M} \lvert a_n \rvert \) to be small, not merely the ordered tail \( \lvert \sum_{n \gt M} a_n \rvert \). For \( \sum (-1)^{n+1}/n \) the ordered tail is small while the absolute tail is infinite — exactly the gap Riemann's theorem exploits.
For the alternating harmonic series construct a rearrangement whose partial sums satisfy \( \liminf_N t_N = -1 \) and \( \limsup_N t_N = +1 \). Describe the first stage numerically and prove the two limit assertions.
Solution

Construction. Apply Step 12 with the constant targets \( \beta_j = 1 \), \( \alpha_j = -1 \): take positive terms \( \tfrac{1}{2k-1} \) in order until the running total first exceeds \( +1 \), then negative terms \( -\tfrac1{2k} \) in order until it first falls below \( -1 \), and repeat forever. Both stages terminate because \( \sum_k \tfrac{1}{2k-1} = \sum_k \tfrac1{2k} = \infty \), and every term is used exactly once, so \( \sigma \) is a bijection.

Stage 1, numerically. Positives: \( 1 \) does not exceed \( 1 \), and \( 1 + \tfrac13 = 1.333333 \gt 1 \), so the first block is \( \{1, \tfrac13\} \) and \( u_1 = 4/3 \); the overshoot is \( u_1 - 1 = 0.333333 = P_2 \), matching the bound of Step 8. Negatives: we need \( \tfrac43 - \tfrac12 H_m \lt -1 \), i.e. \( H_m \gt 14/3 = 4.666667 \). Since \( H_{59} \approx 4.663204 \lt 4.666667 \lt 4.679870 \approx H_{60} \), the least such \( m \) is \( m = 60 \): the block is \( -\tfrac12, -\tfrac14, \dots, -\tfrac1{120} \), sixty terms, ending at \( v_1 = \tfrac43 - \tfrac12 H_{60} = -1.006602 \). The undershoot \( 0.006602 \) is indeed at most \( Q_{60} = \tfrac1{120} = 0.008333 \).

Limits. Let \( u_j \) and \( v_j \) be the top and bottom turning values of stage \( j \). By the minimality rule, \( 1 \lt u_j \le 1 + P_{m_j} \) and \( -1 - Q_{k_j} \le v_j \lt -1 \), and \( P_{m_j}, Q_{k_j} \to 0 \) since \( m_j, k_j \to \infty \) and both reserves are null. Hence \( u_j \to 1 \) and \( v_j \to -1 \), so \( \limsup t_N \ge 1 \) and \( \liminf t_N \le -1 \). Conversely every partial sum lies between two consecutive turning values (the partial sums are monotone inside each block), so \( \min_{i \le j}(v_i) \le t_N \le \max_{i \le j}(u_i) \) for \( N \) in stage \( j \); since \( u_j \le 1 + P_{m_j} \) and \( v_j \ge -1 - Q_{k_j} \) with both corrections tending to \( 0 \), every subsequential limit lies in \( [-1, 1] \). Therefore \( \limsup t_N = 1 \) and \( \liminf t_N = -1 \), and in particular the rearranged series diverges by oscillation even though the original converges to \( \ln 2 \).