Tutorials › Real Analysis › The Riemann Rearrangement Theorem

Home  /  Tutorials  /  Infinite Series

Infinite Series · Tutorial 537 of 1000

The Riemann Rearrangement Theorem

Learn how to construct a rearrangement of a conditionally convergent series whose partial sums approach a chosen real target.

Advanced 11 min read

What You'll Learn

  • Identify why the positive and negative terms of a conditionally convergent series each have infinite total magnitude
  • Construct finite blocks that cross a prescribed target from alternating sides
  • Bound each block’s overshoot by its final term
  • Prove that the constructed ordering is a permutation and its partial sums converge to the target
  • Handle zero terms without changing the convergence argument
  • Apply the construction to several conditionally convergent series

Rearranging Terms to Reach a Chosen Sum

For an absolutely convergent series, rearrangement invariance shows that changing the order cannot change the sum. Conditional convergence permits a very different outcome: by choosing the order carefully, the partial sums can be made to approach any prescribed real number. The Riemann Rearrangement Theorem, stated in Conditional Convergence, describes this phenomenon. Here we develop its block construction and prove that the resulting ordering really is a rearrangement with the desired sum.

The construction uses the positive terms to move a partial sum above the target, then the negative terms to move it below, and repeats. Each move is made by adding a finite block and stopping at the first strict crossing. The amount by which a block passes the target is therefore controlled by its last term. Since the terms tend to zero, these crossings become increasingly accurate.

Positive and Negative Terms

Let \(\sum_{n=1}^{\infty}a_n\) be conditionally convergent. List its strictly positive terms, in their original order, as \(p_1,p_2,\ldots\), and write the magnitudes of its strictly negative terms, also in their original order, as \(q_1,q_2,\ldots\). Thus each \(p_j>0\), and the corresponding negative terms are \(-q_j\), with \(q_j>0\).

The Positive and Negative Parts Theorem from Conditional Convergence gives the key facts: there are infinitely many terms of each sign, and

$$ \sum_{j=1}^{\infty}p_j=\infty, \qquad \sum_{j=1}^{\infty}q_j=\infty. $$

Also, the Necessary Condition for Series Convergence gives \(a_n\to0\). Consequently, \(p_j\to0\) and \(q_j\to0\): each list is made from terms of the original sequence, and its indices tend to infinity. These facts ensure both that a finite block can always cross the target and that the overshoots eventually shrink.

Definition: Given a target \(x\in\mathbb{R}\), a positive crossing block starts at a sum \(s\leq x\) and adds consecutive unused positive terms until the sum is strictly greater than \(x\). A negative crossing block starts at a sum \(s\geq x\) and subtracts consecutive unused magnitudes \(q_j\) until the sum is strictly less than \(x\).

Finite Crossings and Their Overshoots

The infinite total of each sign is useful even after finitely many terms have already been used. Removing a finite initial segment from a positive series whose partial sums are unbounded leaves partial sums that are still unbounded. Thus, from any starting sum at or below \(x\), a finite number of the remaining positive terms must raise the sum strictly above \(x\). The negative case follows in the same way using the divergent sum of the \(q_j\).

Lemma (Crossing-Block Estimate): A positive crossing block starting at \(s\leq x\) is finite, and its endpoint \(t\) satisfies $$ 0<t-x\leq p_r, $$ where \(p_r\) is the last term in the block. A negative crossing block starting at \(s\geq x\) is finite, and its endpoint \(u\) satisfies $$ 0<x-u\leq q_r, $$ where \(q_r\) is the last magnitude used.

Proof. For a positive block, let \(r\) be the first number of unused positive terms for which their sum, when added to \(s\), is strictly greater than \(x\). Such an \(r\) exists by divergence of the remaining positive series. Before the last term is added, the partial sum is at most \(x\), by the minimality of \(r\). Write that preceding sum as \(v\). The endpoint is \(t=v+p_r\), and \(v\leq x<t\). Therefore

$$ 0<t-x=v+p_r-x\leq p_r, $$

because \(v-x\leq0\). This also proves that the endpoint is strictly above \(x\). For a negative block, choose the first \(r\) for which subtracting the next \(r\) unused magnitudes takes the sum strictly below \(x\). The block is finite because the remaining \(q_j\) have infinite total. If \(v\geq x\) is the sum just before its last subtraction, then \(u=v-q_r<x\), so

$$ 0<x-u=x-v+q_r\leq q_r. $$

The inequality follows from \(x-v\leq0\). This proves both estimates. \(\square\)

The starting sum in a block need not be strictly on the appropriate side of the target. In particular, it may equal \(x\). The proof still applies: the first positive term, for example, may already cross strictly above \(x\). The rule is to stop at the first strict crossing, not to add an extra term after a crossing has occurred.

The Riemann Rearrangement Theorem

Theorem (Riemann Rearrangement Theorem): If \(\sum_{n=1}^{\infty}a_n\) is conditionally convergent, then for every \(x\in\mathbb{R}\) there is a permutation \(\pi\) of the positive integers such that $$ \sum_{k=1}^{\infty}a_{\pi(k)}=x. $$

Proof. Fix \(x\in\mathbb{R}\). Start with a running sum of \(0\). If \(0\leq x\), first take a positive crossing block; if \(0>x\), first take a negative crossing block. Thereafter alternate positive and negative crossing blocks. Thus every positive block ends strictly above \(x\), and every negative block ends strictly below \(x\). Each block is finite by the Crossing-Block Estimate. Its terms are taken in their original order from the list of positive terms or from the list of negative terms, always using the next unused term of that sign.

This procedure uses every positive and negative term. There are infinitely many blocks, since the procedure continues indefinitely and every block is finite. The signs alternate after the first block, so there are infinitely many positive blocks and infinitely many negative blocks. Each such block uses at least one new term of its sign. Since the terms of each sign are used in their original order, after sufficiently many blocks every fixed positive term and every fixed negative term has been used.

It remains to account for zero terms, if there are any. If there are finitely many, place all of them at the beginning of the ordering; this changes no partial sum. If there are infinitely many, insert the next unused zero term after each completed positive-negative pair of blocks. In either case, every zero term is included. The ordering so obtained contains every original index exactly once: no nonzero term is repeated, all nonzero terms are eventually used, and all zero terms are included. It therefore defines a permutation \(\pi\) of the positive integers.

We now prove convergence to \(x\), including for partial sums that stop inside a block. At the endpoint of a positive block, the excess above \(x\) is at most the last positive term in that block. At the endpoint of a negative block, the deficit below \(x\) is at most the last magnitude used in that block. As the block number increases, the indices of the last terms used tend to infinity: each block consumes a new term, and only finitely many original terms have index below any fixed bound. Since \(a_n\to0\), the magnitudes of these last terms tend to zero. Hence the block endpoints tend to \(x\).

Within a positive block the partial sums increase, so every intermediate sum lies between the block’s starting and ending sums. Within a negative block they decrease and likewise remain between those endpoints. For any \(\varepsilon>0\), all sufficiently late block endpoints are within \(\varepsilon\) of \(x\). Therefore every partial sum inside any sufficiently late block is also within \(\varepsilon\) of \(x\). Inserted zero terms leave the sum unchanged. The rearranged partial sums thus converge to \(x\), proving the theorem. \(\square\)

Worked Applications

Worked Example: Targeting Zero with an Alternating Harmonic Series

Consider the conditionally convergent series

$$ 1-1+\frac12-\frac12+\frac13-\frac13+\cdots. $$

Its partial sums after each pair are \(0\), while the partial sum after the next positive term is \(1/(n+1)\); both types tend to \(0\). The positive terms are \(p_j=1/j\), and the negative magnitudes are \(q_j=1/j\). Both lists have divergent sums, so the construction applies with target \(x=0\).

The first positive block stops immediately: adding \(p_1=1\) takes the sum from \(0\) to \(1>0\). It does not include \(p_2\). In the first negative block, subtracting \(q_1=1\) takes the sum from \(1\) to \(0\), which is not yet strictly below the target. Subtracting \(q_2=1/2\) then gives \(-1/2<0\). The next positive block begins at \(-1/2\): adding \(p_2=1/2\) reaches \(0\), and adding \(p_3=1/3\) crosses to \(1/3\). The procedure continues, with each block stopping at its first strict crossing. Its overshoot is no greater than its final term, and those terms tend to zero.

Worked Example: The First Crossing for an Inverse-Square-Root Series

Take \(a_n=(-1)^{n-1}/\sqrt{n}\). The Alternating Series Test gives convergence, and the \(p\)-Series Criterion gives divergence of \(\sum 1/\sqrt n\), so the series is conditionally convergent. Its positive terms begin \(1,1/\sqrt3,\ldots\), and its negative magnitudes begin \(1/\sqrt2,1/2,\ldots\). Choose target \(x=0\).

The first positive block begins at \(0=x\). The first term \(a_1=1\) already makes the sum strictly positive, so the block ends there, with endpoint \(1\). The first negative term changes the sum to \(1-1/\sqrt2\), which is still positive because \(1/\sqrt2<1\). The next negative term gives

$$ 1-\frac{1}{\sqrt2}-\frac12<0, $$

since \(1/\sqrt2>1/2\). Thus the negative block ends at this first strict crossing. Its distance below zero is at most the magnitude \(1/2\) of its last term, as the crossing estimate guarantees. Adding \(1/\sqrt3\) to this endpoint makes the sum positive: indeed, \(1/\sqrt3>1/2\), while \(1-1/\sqrt2<1/2\), so

$$ 1-\frac{1}{\sqrt2}-\frac12+\frac{1}{\sqrt3} > -\frac12+\frac{1}{\sqrt3} >0. $$

This example illustrates why stopping at the first strict crossing matters. Adding a second positive term to the initial block would not follow the construction’s rule and would invalidate its stated bound on the first overshoot.

Worked Example: A Negative Target with Infinitely Many Zero Terms

Define \(c_{3j-2}=1/j\), \(c_{3j-1}=0\), and \(c_{3j}=-1/j\) for \(j\geq1\). The original series converges to \(0\): each complete triple sums to zero, and the partial sum after the first term of a triple is \(1/j\), which tends to zero. It is not absolutely convergent, since its absolute values include the divergent harmonic terms twice.

Set the target to \(x=-1\). The initial sum \(0\) is above \(x\), so begin with a negative block. Subtracting \(q_1=1\) reaches \(-1\), not strictly below it; subtracting \(q_2=1/2\) gives \(-3/2<-1\). The next positive block starts at \(-3/2\). Adding \(p_1=1\) gives \(-1/2>-1\), so this block ends there. The next negative block starts at \(-1/2\): subtract \(q_3=1/3\) to reach \(-5/6\), then subtract \(q_4=1/4\) to reach \(-13/12<-1\). The theorem’s construction continues in the same way. Insert one unused zero after each positive-negative pair of blocks; these zero terms do not change any partial sum, and all are eventually included.

What the Construction Does—and Does Not—Say

The size of a crossing block is not required to be uniformly bounded. A block may contain many terms before it reaches the target, especially when the remaining terms of the needed sign are small. The argument needs only that each block be finite and that its overshoot be controlled by its last term. Divergence of the positive and negative parts guarantees the first property; the fact that the terms tend to zero guarantees the second becomes increasingly effective.

It is also important not to confuse a target sum with the sum of the original series. The constructed rearrangement still contains exactly the original terms, but conditional convergence means the order can change the limit. For an absolutely convergent series, the rearrangement invariance results established earlier rule out this behavior. In the conditional case, the divergent positive and negative parts provide the freedom needed to cross any finite target repeatedly.

Takeaway: For a conditionally convergent series, alternate finite blocks of unused positive and negative terms, stopping each block at its first strict crossing of the target. Each overshoot is bounded by the last term used, so the block endpoints—and every intermediate partial sum—approach the chosen real number.

Check Your Understanding

Use the crossing-block construction and its estimates to answer the following questions.

  1. Why do the positive terms and the magnitudes of the negative terms each have infinite total for a conditionally convergent series?
  2. Why does a positive block still cross in finitely many terms when its starting sum is exactly the target?
  3. At a positive crossing, why is the excess above the target no greater than the last positive term?
  4. How does the construction ensure that every positive and negative term is eventually used?
  5. How can infinitely many zero terms be included without affecting convergence to the target?

Previous

Next

Back to Tutorials