3.6 The Riemann rearrangement theorem

This section is non-examinable

Absolutely convergent and conditionally convergent series have very different properties. To illustrate this, we shall discuss what is perhaps one of the most surprising results in undergraduate mathematics: the Riemann rearrangement theorem.

We first consider rearrangements of (finite) sums of real numbers. Addition of real numbers is commutative. If we pick $5$ numbers $a_{1},\dots,a_{5}\in\mathbb{R}$ , say, then the value of their sum does not depend on the order in which we carry out the addition. For instance,

(3.13) (3.13)

a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=a_{2}+a_{5}+a_{4}+a_{1}+a_{3}.

We express this property as follows: given any bijection $\sigma\colon\{1,2,3,4,5\}\to\{1,2,3,4,5\}$ , we have

a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=a_{\sigma(1)}+a_{\sigma(2)}+a_{\sigma(3)}+a_{% \sigma(4)}+a_{\sigma(5)}.

Here the bijection corresponds to a choice of order in which to carry out the sum; for instance, in (3.13) we choose $\sigma$ so that $\sigma(1)=2$ , $\sigma(2)=5$ , $\sigma(3)=4$ , $\sigma(4)=1$ , $\sigma(5)=3$ . Of course, there is nothing special about taking $5$ numbers here, and it is equally true that given $a_{1},\dots,a_{n}\in\mathbb{R}$ for $n\in\mathbb{N}$ we have

(3.14) (3.14)

\sum_{k=1}^{n}a_{k}=\sum_{i=1}^{n}a_{\sigma(i)}\qquad\text{for any bijection $% \sigma\colon\{1,\dots,n\}\to\{1,\dots,n\}$.}

We refer to the sum on the right-hand side of (3.14) as a rearrangement of the sum of the left.

Suppose we now have a sequence of real numbers $(a_{k})_{k\in\mathbb{N}}$ such that $\sum_{k=1}^{\infty}a_{k}$ converges. Given a bijection $\sigma\colon\mathbb{N}\to\mathbb{N}$ we can consider a rearrangement $\sum_{i=1}^{\infty}a_{\sigma(i)}$ of the series. We might naively expect an analogous formula to (3.14) to hold in this context. However, it turns out that this fails dramatically for conditionally convergent series!

Theorem 3.56 (Riemann rearrangement theorem).

Let $(a_{k})_{k\in\mathbb{N}}$ be a sequence of real numbers and suppose $\sum_{k=1}^{\infty}a_{k}$ is conditionally convergent. For any $s\in\mathbb{R}$ , there exists a bijection $\sigma\colon\mathbb{N}\to\mathbb{N}$ such that the rearrangement $\sum_{i=1}^{\infty}a_{\sigma(i)}$ converges to $s$ .

It is worth pausing to appreciate how surprising and counter-intuitive Theorem 3.56 is. We emphasise that we are completely free to choose $s\in\mathbb{R}$ . Thus, the theorem tells us that by rearranging the sum we can change the value of the limit! Remember that a rearrangement $\sum_{k=1}^{\infty}a_{\sigma(k)}$ is simply formed by taking the terms in the partial sums in a different order. For example, we might take

a_{3}+a_{7}+a_{1}+a_{100}+a_{2}+\dots.

The fact that $\sigma\colon\mathbb{N}\to\mathbb{N}$ is a bijection ensures that we do not miss any term in the series and no term appears more than once.

To see how this is this possible, below we shall run through the proof of a special case of Theorem 3.56. The example in fact already contains all the essential ideas, and the method used can be adapted to prove the whole theorem.

Figure 3.4: A schematic for the proof of the Riemann rearrangement theorem. The plot shows the values of the partial sums of the rearrangement

s_{n}:=\sum_{i=1}^{n}a_{\sigma(i)}

. We successively choose

a_{\sigma(1)}

a_{\sigma(2)}

a_{\sigma(3)}>0

and then

a_{\sigma(4)}

a_{\sigma(5)}

a_{\sigma(6)}<0

and so on, in order to home in on the target limit value

s

Example 3.57.

Take $a_{k}:=(-1)^{k}/k$ for $k\in\mathbb{N}$ , so that we have the series $\sum_{k=1}^{\infty}\frac{(-1)^{k}}{k}$ from earlier. We know this series is conditionally convergent, so Theorem 3.56 applies. We now choose some value of $s$ . Let’s take $s:=1868$ , since this corresponds to the year Riemann published his work on the rearrangement theorem. Then by Theorem 3.56 there exists some bijection $\sigma\colon\mathbb{N}\to\mathbb{N}$ such that

\sum_{i=1}^{\infty}a_{\sigma(i)}=1868.

Alternatively, for some personalised mathematics we could take $s$ to be your birthday and find another bijection such that the rearranged sequence converges to that value.

Here we describe explicitly how to construct such a bijection $\sigma\colon\mathbb{N}\to\mathbb{N}$ , and thereby prove Theorem 3.56 in this case.

We first consider separately the sequences of (positive) even terms $(a_{2k})_{k\in\mathbb{N}}$ and (negative) odd terms $(a_{2k+1})_{k\in\mathbb{N}_{0}}$ .

(a) The partial sums of the corresponding sequence of even terms are given by

e_{n}:=\sum_{k=1}^{n}a_{2k}=\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\sum_{k=1}^{% n}\frac{1}{k}.

In particular, up to multiplying by $1/2$ , they agree with the partial sums of the harmonic series. We therefore know that the sequence $(e_{n})_{n\in\mathbb{N}}$ is unbounded and satisfies $e_{n}\to\infty$ as $n\to\infty$ .

(b) The partial sums corresponding to the sequence of odd terms are given by

o_{n}:=\sum_{k=1}^{n}\frac{-1}{2k-1}=-\sum_{k=1}^{n}\frac{1}{2k-1}\leq-\frac{1% }{2}\sum_{k=1}^{n}\frac{1}{k},

where we have used the inequality $2k-1\leq 2k$ . In particular, $(o_{n})_{n\in\mathbb{N}}$ is also unbounded and satisfies $o_{n}\to-\infty$ as $n\to\infty$ .

We now carry out a sequence of steps to construct the rearrangement, as illustrated for an example in Figure 3.4:

•

We begin by adding together (positive) even terms of the series until we pass the threshold $1868$ . More precisely, since $e_{n}\to\infty$ as $n\to\infty$ , we can find some least value $M_{1}\in\mathbb{N}$ such that

(3.15) (3.15) $\sum_{k=1}^{M_{1}}a_{2k}>1868.$
•

We have now overshot our desired limit value. To correct for this, we add together (negative) odd terms until we again drop below the threshold. More precisely, since $o_{n}\to-\infty$ , there exists a least value $N_{1}\in\mathbb{N}$ such that

(3.16) (3.16) $\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}}a_{2k-1}<1868.$
•

We have now undershot our desired limit value. To correct for this, we again add together (positive) even terms until we again rise above the threshold. In particular, we let $M_{2}\in\mathbb{N}$ be the least value such that

$\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}}a_{2k-1}+\sum_{k=M_{1}+1}^{M_{2}}a_% {2k}>1868.$
•

Repeating this process of overshooting and undershooting the desired limit, we recursively define increasing sequences of integers $M_{1}<M_{2}<\cdots$ and $N_{1}<N_{2}\cdots$ such that

$\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}}a_{2k-1}+\cdots+\sum_{k=M_{J-1}+1}^% {M_{J}}a_{2k}>1868$

and

$\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}}a_{2k-1}+\cdots+\sum_{k=M_{J-1}+1}^% {M_{J}}a_{2k}+\sum_{k=N_{J-1}+1}^{N_{J}}a_{2k-1}<1868$

for each $J\in\mathbb{N}$ . This defines a rearrangement $\sum_{i=1}^{\infty}a_{\sigma(i)}$ of the series. Indeed, it is clear that every term will appear in our reordering of the sequence, and each term appears at most once.

It remains to show the rearranged series converges to the target value $1868$ . The intuitive idea is very simple: since $a_{n}\to 0$ as $n\to\infty$ , the amount we overshoot or undershoot the target in each step must tend to $0$ . However, writing out the details is a little involved, so we only provide a sketch.

Let’s go back to the first step and try to measure the extent to which we overshoot the limit at the beginning of the construction. Since $M_{1}\in\mathbb{N}$ is the least value satisfying (3.15), we must have

\sum_{k=1}^{M_{1}-1}a_{2k}\leq 1868\qquad\text{and so}\qquad 1868<\sum_{k=1}^{% M_{1}}a_{2k}\leq 1868+a_{2M_{1}}.

Thus, the amount by which we overshoot the limit is bounded by $a_{2M_{1}}$ .

Similarly, in the second step we chose $N_{1}$ to be the least value such that (3.16) holds. This means that

\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}-1}a_{2k-1}\geq 1868.

Moreover, taking into account the signs of the terms and our observations about the first step, we have

1868+a_{2N_{1}-1}\leq\sum_{k=1}^{M_{1}}a_{2k}+\sum_{k=1}^{N_{1}}a_{2k-1}\leq% \sum_{k=1}^{M_{1}}a_{2k}\leq 1868+a_{2M_{1}}.

This means that, at the second step, the sum undershoots 1868 by at most $|a_{2N_{1}-1}|$ and overshoots 1868 by at most $a_{2M_{1}}$ .

Continuing in this way, in later steps of the construction we will either overshoot $1868$ by at most $a_{2M_{J}}$ or undershoot $1868$ by at most $|a_{2N_{J}-1}|$ . Since these values are tending to $0$ as $J\to\infty$ , we conclude that $\sum_{i=1}^{\infty}a_{\sigma(i)}=1868$ holds.

Exercise 3.58.

Let $a_{k}:=(-1)^{k}/k$ for all $k\in\mathbb{N}$ . Explain why there exists a rearrangement $\sum_{i=1}^{\infty}a_{\sigma(i)}$ of the series $\sum_{k=1}^{\infty}a_{k}$ such that $\sum_{i=1}^{n}a_{\sigma(i)}\to\infty$ as $n\to\infty$ (see Definition 2.53). Here you only need to give a reasonably convincing sketch (similar to the example discussed above), rather than spelling out all the technical details.

Remark 3.59.

The Riemann rearrangement theorem applies only to conditionally convergent series. For absolutely convergent series, the situation is completely different. Indeed, suppose $(a_{k})_{k\in\mathbb{N}}$ is a sequence of real numbers and that the series $\sum_{k=1}^{\infty}a_{k}$ is absolutely convergent. Then for any bijection $\sigma\colon\mathbb{N}\to\mathbb{N}$ the rearrangement $\sum_{i=1}^{\infty}a_{\sigma(i)}$ is convergent and, furthermore,

\sum_{k=1}^{\infty}a_{k}=\sum_{i=1}^{\infty}a_{\sigma(i)}.

We will not prove this result in the course, but the details are not too difficult. If you are interested, feel free to ask the lecturer!