# Riemann series theorem

In mathematics, the Riemann series theorem (also called the Riemann rearrangement theorem), named after 19th-century German mathematician Bernhard Riemann, says that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, or diverges.

As an example, the series 1 – 1 + 1/2 – 1/2 + 1/3 – 1/3 + ... converges to 0 (for a sufficiently large number of terms, the partial sum gets arbitrarily near to 0); but replacing all terms with their absolute values gives 1 + 1 + 1/2 + 1/2 + 1/3 + 1/3 + ... , which sums to infinity. Thus the original series is conditionally convergent, and can be rearranged (by taking the first two positive terms followed by the first negative term, followed by the next two positive terms and then the next negative term, etc.) to give a series that converges to a different sum: 1 + 1/2 – 1 + 1/3 + 1/4 – 1/2 + ... = ln 2. More generally, using this procedure with p positives followed by q negatives gives the sum ln(p/q). Other rearrangements give other finite sums or do not converge to any sum.

## Definitions

A series $\sum _{n=1}^{\infty }a_{n}$ converges if there exists a value $\ell$ such that the sequence of the partial sums

$(S_{1},S_{2},S_{3},\ldots ),\quad S_{n}=\sum _{k=1}^{n}a_{k},$ converges to $\ell$ . That is, for any ε > 0, there exists an integer N such that if n ≥ N, then

$\left\vert S_{n}-\ell \right\vert \leq \epsilon .$ A series converges conditionally if the series $\sum _{n=1}^{\infty }a_{n}$ converges but the series $\sum _{n=1}^{\infty }\left\vert a_{n}\right\vert$ diverges.

A permutation is simply a bijection from the set of positive integers to itself. This means that if $\sigma$ is a permutation, then for any positive integer $b,$ there exists exactly one positive integer $a$ such that $\sigma (a)=b.$ In particular, if $x\neq y$ , then $\sigma (x)\neq \sigma (y)$ .

## Statement of the theorem

Suppose that $(a_{1},a_{2},a_{3},\ldots )$ is a sequence of real numbers, and that $\textstyle \sum _{n=1}^{\infty }a_{n}$ is conditionally convergent. Let $M$ be a real number. Then there exists a permutation $\sigma$ such that

$\sum _{n=1}^{\infty }a_{\sigma (n)}=M.$ There also exists a permutation $\sigma$ such that

$\sum _{n=1}^{\infty }a_{\sigma (n)}=\infty .$ The sum can also be rearranged to diverge to $-\infty$ or to fail to approach any limit, finite or infinite.

## Alternating harmonic series

### Changing the sum

The alternating harmonic series is a classic example of a conditionally convergent series:

$\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}$ is convergent, while

$\sum _{n=1}^{\infty }{\bigg |}{\frac {(-1)^{n+1}}{n}}{\bigg |}$ is the ordinary harmonic series, which diverges. Although in standard presentation the alternating harmonic series converges to ln(2), its terms can be arranged to converge to any number, or even to diverge. One instance of this is as follows. Begin with the series written in the usual order,

$1-{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{4}}+\cdots$ and rearrange the terms:

$1-{\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{3}}-{\frac {1}{6}}-{\frac {1}{8}}+{\frac {1}{5}}-{\frac {1}{10}}-{\frac {1}{12}}+\cdots$ where the pattern is: the first two terms are 1 and −1/2, whose sum is 1/2. The next term is −1/4. The next two terms are 1/3 and −1/6, whose sum is 1/6. The next term is −1/8. The next two terms are 1/5 and −1/10, whose sum is 1/10. In general, the sum is composed of blocks of three:

${\frac {1}{2k-1}}-{\frac {1}{2(2k-1)}}-{\frac {1}{4k}},\quad k=1,2,\dots .$ This is indeed a rearrangement of the alternating harmonic series: every odd integer occurs once positively, and the even integers occur once each, negatively (half of them as multiples of 4, the other half as twice odd integers). Since

${\frac {1}{2k-1}}-{\frac {1}{2(2k-1)}}={\frac {1}{2(2k-1)}},$ this series can in fact be written:

${\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{6}}-{\frac {1}{8}}+{\frac {1}{10}}+\cdots +{\frac {1}{2(2k-1)}}-{\frac {1}{2(2k)}}+\cdots$ $={\frac {1}{2}}\left(1-{\frac {1}{2}}+{\frac {1}{3}}-\cdots \right)={\frac {1}{2}}\ln(2)$ which is half the usual sum.

### Getting an arbitrary sum

An efficient way to recover and generalize the result of the previous section is to use the fact that

$1+{1 \over 2}+{1 \over 3}+\cdots +{1 \over n}=\gamma +\ln n+o(1),$ where γ is the Euler–Mascheroni constant, and where the notation o(1) denotes a quantity that depends upon the current variable (here, the variable is n) in such a way that this quantity goes to 0 when the variable tends to infinity.

It follows that the sum of q even terms satisfies

${1 \over 2}+{1 \over 4}+{1 \over 6}+\cdots +{1 \over 2q}={1 \over 2}\,\gamma +{1 \over 2}\ln q+o(1),$ and by taking the difference, one sees that the sum of p odd terms satisfies

${1}+{1 \over 3}+{1 \over 5}+\cdots +{1 \over 2p-1}={1 \over 2}\,\gamma +{1 \over 2}\ln p+\ln 2+o(1).$ Suppose that two positive integers a and b are given, and that a rearrangement of the alternating harmonic series is formed by taking, in order, a positive terms from the alternating harmonic series, followed by b negative terms, and repeating this pattern at infinity (the alternating series itself corresponds to a = b = 1, the example in the preceding section corresponds to a = 1, b = 2):

${1}+{1 \over 3}+\cdots +{1 \over 2a-1}-{1 \over 2}-{1 \over 4}-\cdots -{1 \over 2b}+{1 \over 2a+1}+\cdots +{1 \over 4a-1}-{1 \over 2b+2}-\cdots$ Then the partial sum of order (a+b)n of this rearranged series contains p = an positive odd terms and q = bn negative even terms, hence

$S_{(a+b)n}={1 \over 2}\ln p+\ln 2-{1 \over 2}\ln q+o(1)={1 \over 2}\ln(a/b)+\ln 2+o(1).$ It follows that the sum of this rearranged series is

${1 \over 2}\ln(a/b)+\ln 2=\ln {\bigl (}2{\sqrt {a/b}}{\bigr )}.$ Suppose now that, more generally, a rearranged series of the alternating harmonic series is organized in such a way that the ratio pn / qn between the number of positive and negative terms in the partial sum of order n tends to a positive limit r. Then, the sum of such a rearrangement will be

$\ln {\bigl (}2{\sqrt {r}}{\bigr )},$ and this explains that any real number x can be obtained as sum of a rearranged series of the alternating harmonic series: it suffices to form a rearrangement for which the limit r is equal to  e2x /  4.

## Proof

### Existence of a rearrangement that sums to any positive real M

For simplicity, this proof assumes first that an ≠ 0 for every n. The general case requires a simple modification, given below. Recall that a conditionally convergent series of real terms has both infinitely many negative terms and infinitely many positive terms. First, define two quantities, $a_{n}^{+}$ and $a_{n}^{-}$ by:

$a_{n}^{+}={\frac {a_{n}+|a_{n}|}{2}},\quad a_{n}^{-}={\frac {a_{n}-|a_{n}|}{2}}.$ That is, the series $\sum _{n=1}^{\infty }a_{n}^{+}$ includes all an positive, with all negative terms replaced by zeroes, and the series $\sum _{n=1}^{\infty }a_{n}^{-}$ includes all an negative, with all positive terms replaced by zeroes. Since $\sum _{n=1}^{\infty }a_{n}$ is conditionally convergent, both the positive and the negative series diverge. Let M be a positive real number. Take, in order, just enough positive terms $a_{n}^{+}$ so that their sum exceeds M. Suppose we require p terms – then the following statement is true:

$\sum _{n=1}^{p-1}a_{n}^{+}\leq M<\sum _{n=1}^{p}a_{n}^{+}.$ This is possible for any M > 0 because the partial sums of $a_{n}^{+}$ tend to $+\infty$ . Discarding the zero terms one may write

$\sum _{n=1}^{p}a_{n}^{+}=a_{\sigma (1)}+\cdots +a_{\sigma (m_{1})},\quad a_{\sigma (j)}>0,\ \ \sigma (1)<\ldots <\sigma (m_{1})=p.$ Now we add just enough negative terms $a_{n}^{-}$ , say q of them, so that the resulting sum is less than M. This is always possible because the partial sums of $a_{n}^{-}$ tend to $-\infty$ . Now we have:

$\sum _{n=1}^{p}a_{n}^{+}+\sum _{n=1}^{q}a_{n}^{-} Again, one may write

$\sum _{n=1}^{p}a_{n}^{+}+\sum _{n=1}^{q}a_{n}^{-}=a_{\sigma (1)}+\cdots +a_{\sigma (m_{1})}+a_{\sigma (m_{1}+1)}+\cdots +a_{\sigma (n_{1})},$ with

$\sigma (m_{1}+1)<\ldots <\sigma (n_{1})=q.$ The map σ is injective, and 1 belongs to the range of σ, either as image of 1 (if a1 > 0), or as image of m1 + 1 (if a1 < 0). Now repeat the process of adding just enough positive terms to exceed M, starting with n = p + 1, and then adding just enough negative terms to be less than M, starting with n = q + 1. Extend σ in an injective manner, in order to cover all terms selected so far, and observe that a2 must have been selected now or before, thus 2 belongs to the range of this extension. The process will have infinitely many such "changes of direction". One eventually obtains a rearrangement  ∑ aσ (n). After the first change of direction, each partial sum of  ∑ aσ (n) differs from M by at most the absolute value $a_{p_{j}}^{+}$ or $|a_{q_{j}}^{-}|$ of the term that appeared at the latest change of direction. But ∑ an converges, so as n tends to infinity, each of an, $a_{p_{j}}^{+}$ and $a_{q_{j}}^{-}$ go to 0. Thus, the partial sums of  ∑ aσ (n) tend to M, so the following is true:

$\sum _{n=1}^{\infty }a_{\sigma (n)}=M.$ The same method can be used to show convergence to M negative or zero.

One can now give a formal inductive definition of the rearrangement σ, that works in general. For every integer k ≥ 0, a finite set Ak of integers and a real number Sk are defined. For every k > 0, the induction defines the value σ(k), the set Ak consists of the values σ(j) for j ≤ k and Sk is the partial sum of the rearranged series. The definition is as follows:

• For k = 0, the induction starts with A0 empty and S0 = 0.
• For every k ≥ 0, there are two cases: if Sk ≤ M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in Ak and an ≥ 0; if Sk > M, then σ(k+1) is the smallest integer n ≥ 1 such that n is not in Ak and an < 0. In both cases one sets
$A_{k+1}=A_{k}\cup \{\sigma (k+1)\}\,;\quad S_{k+1}=S_{k}+a_{\sigma (k+1)}.$ It can be proved, using the reasonings above, that σ is a permutation of the integers and that the permuted series converges to the given real number M.

### Existence of a rearrangement that diverges to infinity

Let $\textstyle \sum _{i=1}^{\infty }a_{i}$ be a conditionally convergent series. The following is a proof that there exists a rearrangement of this series that tends to $\infty$ (a similar argument can be used to show that $-\infty$ can also be attained).

Let $p_{1} be the sequence of indexes such that each $a_{p_{i}}$ is positive, and define $n_{1} to be the indexes such that each $a_{n_{i}}$ is negative (again assuming that $a_{i}$ is never 0). Each natural number will appear in exactly one of the sequences $(p_{i})$ and $(n_{i}).$ Let $b_{1}$ be the smallest natural number such that

$\sum _{i=1}^{b_{1}}a_{p_{i}}\geq |a_{n_{1}}|+1.$ Such a value must exist since $(a_{p_{i}}),$ the subsequence of positive terms of $(a_{i}),$ diverges. Similarly, let $b_{2}$ be the smallest natural number such that:

$\sum _{i=b_{1}+1}^{b_{2}}a_{p_{i}}\geq |a_{n_{2}}|+1,$ and so on. This leads to the permutation

$(\sigma (1),\sigma (2),\sigma (3),\ldots )=(p_{1},p_{2},\ldots ,p_{b_{1}},n_{1},p_{b_{1}+1},p_{b_{1}+2},\ldots ,p_{b_{2}},n_{2},\ldots ).$ And the rearranged series, $\textstyle \sum _{i=1}^{\infty }a_{\sigma (i)},$ then diverges to $\infty$ .

From the way the $b_{i}$ were chosen, it follows that the sum of the first $b_{1}+1$ terms of the rearranged series is at least 1 and that no partial sum in this group is less than 0. Likewise, the sum of the next $b_{2}-b_{1}+1$ terms is also at least 1, and no partial sum in this group is less than 0 either. Continuing, this suffices to prove that this rearranged sum does indeed tend to $\infty .$ ### Existence of a rearrangement that fails to approach any limit, finite or infinite

In fact, if $\sum _{n=1}^{\infty }a_{n}}$ is conditionally convergent, then there is a rearrangement of it such that the partial sums of the rearranged series form a dense subset of $\mathbb {R}$ .[citation needed]

## Generalizations

### Sierpiński theorem

In Riemann's theorem, the permutation used for rearranging a conditionally convergent series to obtain a given value in $\mathbf {R} \cup \{\infty ,-\infty \}$ may have arbitrarily many non-fixed points, i.e. all the indexes of the terms of the series may be rearranged. One may ask if it is possible to rearrange only the indexes in a smaller set so that a conditionally convergent series converges to an arbitrarily chosen real number or diverges to (positive or negative) infinity. The answer of this question is positive: Sierpiński proved that is sufficient to rearrange only some strictly positive terms or only some strictly negative terms.

This question has also been explored using the notion of ideals: for instance, Wilczyński proved that is sufficient to rearrange only the indexes in the ideal of sets of asymptotic density zero. Filipów and Szuca proved that other ideals also have this property.

### Steinitz's theorem

Given a converging series ∑ an of complex numbers, several cases can occur when considering the set of possible sums for all series ∑ aσ (n) obtained by rearranging (permuting) the terms of that series:

• the series ∑ an may converge unconditionally; then, all rearranged series converge, and have the same sum: the set of sums of the rearranged series reduces to one point;
• the series ∑ an may fail to converge unconditionally; if S denotes the set of sums of those rearranged series that converge, then, either the set S is a line L in the complex plane C, of the form
$L=\{a+tb:t\in \mathbf {R} \},\quad a,b\in \mathbf {C} ,\ b\neq 0,$ or the set S is the whole complex plane C.

More generally, given a converging series of vectors in a finite-dimensional real vector space E, the set of sums of converging rearranged series is an affine subspace of E.