414/514 The theorems of Riemann and Sierpiński on rearrangement of series

Perhaps the first significant observation in the theory of infinite series is that there are convergent series whose terms can be rearranged to form a new series that converges to a different value.

A well known example is provided by the alternating harmonic series,

$\displaystyle 1-\frac12 +\frac13-\frac14+\frac15-\frac16+\frac17-\dots$

and its rearrangement

$\displaystyle 1-\frac12-\frac14+\frac13-\frac16-\frac18+\frac15-\dots$

According to

Henry Parker Manning. Irrational numbers and their representation by sequences and series. John Wiley & Sons, 1906,

Laurent evaluated the latter by inserting parentheses (see pages 97, 98):

$\displaystyle \left(1-\frac12\right)-\frac14+\left(\frac13-\frac16\right)-\frac18+\left(\frac15-\frac1{10}\right)-\dots$ $\displaystyle=\frac12\left(1-\frac12+\frac13-\frac14+\dots\right)$

A similar argument is possible with the rearrangement

$\displaystyle 1+\frac13-\frac12+\frac15+\frac17-\frac14+\dots$ ,

which can be rewritten as

$\displaystyle 1+0+\frac13-\frac12+\frac15+0+\frac17+\dots$ $\displaystyle =\left(1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\dots\right)$ $\displaystyle +\left(0+\frac12+0-\frac14+0+\frac16+0-\dots\right)$ $\displaystyle =\frac32\left(1-\frac12+\frac13-\frac14+\dots\right).$

The first person to realize that rearranging the terms of a series may change its sum was Dirichlet in 1827, while working on the convergence of Fourier series. (The date is mentioned by Riemann in his Habilitationsschrift, see also page 94 of Ivor Grattan-Guinness. The Development of the Foundations of Mathematical Analysis from Euler to Riemann. MIT, 1970.)

Ten years later, he published

G. Lejeune Dirichlet. Beweis des Satzes, dass jede unbegrenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftlichen Factor sind, unendlich viele Primzahlen enthält. Abhandlungen der Königlich Preussischen Akademie der Wissenschaften von 1837, 45-81,

where he shows that this behavior is exclusive of conditionally convergent series:

Theorem (Dirichlet). If a series converges absolutely, all its rearrangements converge to the same value.

Proof. Let $u_0,u_1,\dots$ be the original sequence and $u_{\pi(0)},u_{\pi(1)},\dots$ a rearrangement. Denote by $U_0,U_1,\dots$ and $V_0,V_1,\dots$ their partial sums, respectively. Fix $\epsilon>0$ . We have that for any $n$ , if $m$ is large enough, then for all $i\le n$ there is some $j\le m$ with $\pi(j)=i$ . Also, there is a $k$ such that for all $j\le m$ there is a $i\le k$ with $\pi(j)=i$ , so

$|U_m-V_m|\le\sum_{i=n+1}^m|u_i|+\sum_{i=n+1}^k|u_j|.$

Choosing $n$ large enough, and using that $\sum_i|u_i|$ converges, we can ensure that the two displayed series add up to less than $\epsilon$ . This gives the result. $\Box$

II.

In 1853, Riemann proved his rearrangement theorem, although it was not published until 1866, as part of his Habilitationsschrift on representation of functions as trigonometric series, Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe. See here for Riemann’s papers.

Theorem (Riemann). Given any $\alpha\le\beta$ in the extended reals, a conditionally convergent series of reals can be rearranged so that the liminf of the partial sums of the rearranged series is $\alpha$ , while the limsup is $\beta$ . In particular, any real can be obtained as the sum of some rearrangement of the original series.

Proof. The idea is very simple, and based on the following observation:

Lemma. If $u_0,u_1,\dots$ is a sequence of real numbers such that $\sum_i u_i$ conditionally converges, then, letting $a_0,a_1,\dots$ and $-b_0,-b_1,\dots$ , respectively, denote the subsequence of positive and of non-positive terms, we have that both sequences are infinite, both converge to zero, and for any $k$ , we have that $\displaystyle \sum_{n=k}^\infty a_n=\sum_{n=k}^\infty b_n=+\infty$ .

Proof. The point is that, up to padding the subsequences with zeroes at appropriate places, $\sum_i |u_i|=\sum_i a_i+\sum_i b_i$ , so at least one of the two series must diverge, since the original series is not absolutely convergent, by assumption. If one of the series converges, then the other, being a difference of two convergent series, would converge as well. It follows that both series diverge (to $+\infty$ ). Since both the $a_i$ and the $-b_i$ are subsequences of $u_i$ , which converges to $0$ (since $\sum_i u_i$ converges), then they converge to zero as well. That, in fact, $\sum_{n=k} a_n$ diverges no matter how large $k$ is, is immediate. $\Box$

Using the lemma, we prove the rearrangement theorem in a straightforward fashion. To explain the idea, suppose first that $\alpha=\beta=r$ is real. Consider the rearrangement that first adds positive terms until the sum is larger than $r$ , then adds non-positive terms until the sum is smaller than $r$ , then adds positive terms again, etc, so the partial sums oscillate being larger and smaller than $r$ , but each time by smaller amounts, since the $a_i$ and the $b_i$ converge to $0$ .

The general case follows the same outline: If $\alpha$ is real, define $\alpha_n=\alpha$ for all $n$ . If $\alpha=+\infty$ , define $\alpha_n=n$ , and if $\alpha=-\infty$ , define $\alpha_n=-n$ . Define the sequence $\beta_n$ similarly.

We define the rearrangement $v_0,v_1,\dots$ by stages, and use $V_0,V_1,\dots$ to denote the partial sums of the rearranged sequence. At the beginning of stage $k$ , we have used an initial segment $a_0,\dots,a_{t_{k-1}}$ of the subsequence of positive terms, and an initial segment $-b_0,-b_1,\dots,-b_{s_{k-1}}$ of the subsequence of non-positive terms, and we have defined $v_0,v_1,\dots,v_{m_{k-1}}$ (where $m_{k-1}=t_{k-1}+s_{k-1}$ ). At stage $k=0$ , we begin with $V_{-1}=0$ and $t_{-1}=s_{-1}=-1$ . We continue the definition by setting $v_{m_{k-1}+1}=a_{t_{k-1}+1}$ , $v_{m_{k-1}+2}=a_{t_{k-1}+2}$ , etc, until an index $n$ is reached such that $V_{m_{k-1}}+(a_{t_{k-1}+1}+\dots+a_n)>\beta_n$ . Note that $n$ exists, since $\sum_{i>t_{k-1}}a_i=+\infty.$ We stop at the least such $n$ , and set $t_k=n$ . Let $l_k=m_{k-1}+(n-t_{k-1})$ .

We continue with $v_{l_k+1}=-b_{s_{k-1}+1}$ , $v_{l_k+2}=-b_{s_{k-1}+2}$ , etc, until an index $n$ is reached such that $V_{l_k}-(b_{s_{k-1}+1}+\dots+b_n)<\alpha_n$ . Again, $n$ exists because $\sum_{i>s_{k-1}}b_i=+\infty$ . The least such $n$ we call $s_k$ , and this concludes stage $k$ .

It is now a routine matter to check that this rearrangement has the desired properties: The sequence of partial sums $V_{l_0}, V_{l_1},\dots$ converges to $\beta$ , because $V_{l_k}-\beta_k\to 0$ , since $a_i\to0$ . Similarly, the sequence $V_{m_0},V_{m_1},\dots$ converges to $\alpha$ . This shows that the limsup of the partial sums is at least $\beta$ , and the liminf is at most $\alpha$ . But the $V_i$ with $m_{k-1}\le i\le V_{l_k}$ are increasing, and the $V_i$ with $l_k\le i\le m_k$ are decreasing, so no larger limit than $\beta$ or smaller than $\alpha$ can be achieved. $\Box$

Corollary. There is an injection of $\mathbb R$ into the set $S_\infty$ of permutations of the natural numbers.

The idea here is that we can begin with our favorite conditionally convergent series, and assign to $r\in\mathbb R$ the rearrangement with series converging to $r$ given by the theorem. Of course there are other methods of establishing the corollary, but I find this argument cute.

Exercise. Modify the construction to show that a conditionally convergent series admits a rearrangement such that any extended real is the limit of a subsequence of the sequence of partial sums.

III.

Prior to Riemann’s theorem, Ohm had obtained interesting related results, that were later extended by Schlömilch, and Pringsheim (see here and here for statements of some of their theorems, and further recent work). They are usually discussed in terms of rearrangements of the alternating harmonic series, although their results are more general. See also these slides by Marion Scheepers for additional work along these lines, and for details see

Marion Scheepers. On the Pringsheim rearrangement theorems. J. Math. Anal. Appl., 267 (2), (2002), 418–433. MR1888013 (2003j:40002).

(In the slides, Scheepers is using the notion of signwise monotonic series. This just means series $\sum_i u_i$ where ${}|u_0|,|u_1|,|u_2|,\dots$ is monotonically decreasing.)

Ohm’s theorem appear in

Martin Ohm. De nonnullis seriebus infinitis summandis Commentatio, Berlin, 1839.

It predicts the value of the series obtained if we fix positive integers $p$ and $q$ , and the terms of the alternating harmonic series are rearranged so that first we add the first $p$ positive terms, then the first $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative ones, etc. For instance, if $p=1$ and $q=2$ , the rearrangement is

$1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + 1/5 - \dots$

Theorem (Ohm). The rearrangement of the alternating harmonic series obtained by adding first the first $p$ positive terms, then the first $q$ negative terms, then the next $p$ positive terms, then the next $q$ negative terms, etc, converges to $\ln(2) + (1/2) \ln(p/q)$ .

In particular, for $p=1, q=2$ , the limit is $(1/2) \ln(2)$ and if $p=2,q=1$ , the limit is $(3/2)\ln(2),$ which coincides with the observations we began with.

(See also here.)

[My Master’s student Monica Agana is working on her thesis on the classical theory of rearrangements, covering in detail the results by Ohm, Schlömilch, and Pringsheim.]

IV.

A rearrangement is sometimes called simple iff the subsequences of positive terms of the original series and of the rearranged series coincide, and similarly with the subsequence of non-positive terms. Note that the rearrangements in Riemann’s and Ohm’s theorems are simple.

Wacław Sierpiński. Sur une propriété des séries qui ne sont pas absolument convergentes, Bull. Intern. Acad. Sci.: Cracovie A (1911) 149–158,

Sierpiński considers a different kind of rearrangements, where the non-positive terms are all fixed, and only the positive terms are permuted. (The paper is available here.) His result is the following:

Theorem (Sierpiński).

Let $\sum_i u_i$ be a conditionally convergent series of real numbers that converges to $U$ . For any $V\le U$ there is a rearrangement $v_0,v_1,\dots$ of the $u_i$ such that $\sum_i v_i=V$ , and the rearrangement leaves fixed all non-positive terms.

Similarly, for any $W\ge U$ there is a rearrangement $w_0,w_1,\dots$ of the $u_i$ such that $\sum_i w_i=W$ , and the rearrangement leaves fixed all positive terms.

Moreover, for any extended real $\alpha$ , there is a rearrangement $s_0,s_1,\dots$ of the $u_i$ such that $\sum_i s_i=\alpha$ and, for every $n$ , $u_n>0$ iff $s_n>0$ .

Before we proceed with the proof, let me mention a related question. Given a conditionally convergent series $\sum_i u_i$ with sum $U$ , it follows from the theorem that if there is a rearrangement $v_0,v_1,\dots$ fixing non-negative terms and such that $\sum_i v_i=V$ , then any numbers in $(-\infty,V]$ can be so obtained, regardless of whether $V\le U$ or not, since a rearrangement of the $v_i$ fixing its non-negative terms is also a rearrangement of the $u_i$ with the same property. Let $s$ be the supremum of the numbers $V$ obtained via such rearrangements.

Question. Can $s=+\infty$ ? If $s\in\mathbb R$ , is $s$ one of these numbers $V$ ?

Proof. Note first that it suffices to prove (1), since (2) is an immediate consequence of it: Simply apply (1) to the series $\sum_i -u_i$ , noting that $-W\le -U$ to obtain (2). For (3), we can use (1) or (2) if $\alpha$ is real. But it is an easy exercise to show (3) if $\alpha=+\infty$ or $\alpha=-\infty$ .

For completeness, I sketch a proof of (3) when $\alpha=+\infty$ . The case $\alpha=-\infty$ follows the same outline, just as (2) follows from (1). Let $a_0,a_1,\dots$ and $-b_0,-b_1,\dots$ denote, respectively, the subsequences of positive and of non-positive terms of the $u_i$ , so $\sum_i a_i=\sum_i b_i=+\infty$ . Split the sequence of $b_i$ into two disjoint subsequences $b_0',b_1',\dots$ and $c_0,c_1,\dots$ such that $\sum_i b_i'=+\infty$ but $\sum_i c_i<+\infty$ , and $c_i\ne 0$ for all $i$ .

We describe the rearrangement required in (3) by stages. Our rearrangement fixes all positive terms and only permutes the negative ones. At the beginning of stage $n$ , the permuted sequence built so far consists precisely of initial segments of the $c_i$ and of the $b_i'$ . In fact, $b_0',b_1',\dots,b_{n-1}'$ have been used, as have, say, $c_0,\dots, c_{k-1}$ .

Continue by using $c_k,c_{k+1},\dots$ whenever a non-positive term should be used in the rearrangement, until at least $c_k$ has been used, and a partial sum is obtained that is strictly larger than

$\displaystyle n+b_n'+\sum c_i.$

(This is possible because $\sum a_i=+\infty$ while $\sum c_i<+\infty$ .) Once this happens, use $-b_n'$ at the first opportunity. This concludes stage $n$ .

By induction one easily checks that all partial sums in stage $n+1$ are larger than $n$ , and therefore the rearranged series diverges to $+\infty$ , as wanted. This completes the proof of (3) for $\alpha=\pm\infty$ .

It remains to verify (1).

Just as Riemann’s theorem depended on a lemma about the series of positive and non-positive terms of the original sequence, Sierpiński’s result depends on a lemma about the series of positive terms. Unlike Riemann’s theorem, in this case, all the work goes into the proof of the lemma, and the theorem is more properly a corollary of it.

Lemma.Suppose $a_0,a_1,\dots$ is a sequence of positive numbers such that $\sum_i a_i=+\infty$ and $\lim_i a_i=0$ Let $A_0,A_1,\dots$ denote the sequence of partial sums. For any $L\ge0$ there is a rearrangement $c_0,c_1,\dots$ with sequence of partial sums $C_0,C_1,\dots$ such that $\lim_i (A_i-C_i)=L$ .

To see that the lemma implies the result, suppose that $V\le U$ , and let $L=U-V$ . As above, let $a_0,a_1,\dots$ and $-b_0,-b_1,\dots$ be, respectively, the subsequences of positive and of non-positive terms of the $u_i$ . Let $A_0,A_1,\dots$ and $B_0,B_1,\dots$ denote, respectively, the sequences of partial sums of the $a_i$ and of the $b_i$ . For each $i$ , let $p_i$ be the number of positive terms among $u_0,\dots,u_i$ , and let $n_i$ be the number of non-positive terms, so that $p_i+n_i=i+1$ , and if $U_0,U_1,\dots$ denote the partial sums of the $u_i$ , then $U_i=A_{p_i}-B_{n_i}$ for all $i$ , and $p_i,n_i\to+\infty$ .

Apply the lemma to the $a_i$ to obtain a rearrangement $c_0,c_1,\dots$ with partial sums $C_0,C_1,\dots$ such that $A_i-C_i\to L$ . Consider the rearrangement $v_0,v_1,\dots$ of the $u_i$ that fixes the $-b_i$ and permutes the $a_i$ as indicated. If $V_0,V_1,\dots$ are the partial sums of the $V_i$ , note that, by design, $V_i=C_{p_i}-B_{n_i}$ , so that $U_i-V_i=A_{p_i}-C_{p_i}\to L$ , and $V_i\to U-L=V$ , as wanted.

Note that the requirement that $L\ge0$ in the statement of the lemma, and therefore that $V\le U$ in (1), cannot be improved in general to cover a larger range of possible values, since the $a_i$ may be decreasing, in which case we have that $A_i\ge C-i$ for all $i$ .

All that remains is to prove the lemma.

Proof. As in Riemann’s theorem, we proceed by stages. Starting with $A_{-1}=C_{-1}=0$ , at stage $n$ we examine $A_{n-1}-C_{n-1}$ to decide the value of $c_n$ . As in that theorem, we want to arrange that the values $A_i-C_i$ increase if smaller than $L$ , and decrease if larger, so that in the limit we obtain the desired value. Actually, we will need to modify slightly this strategy. To motivate the proof, consider first the case where the $a_i$ are monotonic, that is, $a_0\ge a_1\ge\dots$ In this case, the strategy works: At stage $n$ , we consider two cases, according to whether $A_{n-1}-C_{n-1}\le L$ or $A_{n-1}-C_{n-1}>L$ :

If $A_{n-1}-C_{n-1}\le L$ , then let $r$ be such that $a_r<a_n/2$ , and set $c_n=a_r$ . This is possible, since $a_r\to 0$ .

Note that, for as long as $A_k-C_k\le L$ , we stay in this case. Thus, if from some point $n$ on we are always in case 1, then for $k$ large enough we have

$\displaystyle A_k-C_k=(A_{n-1}-C_{n-1})+\sum_{i=n}^k (a_i-c_i)\ge(A_{n-1}-C_{n-1})+$ $\displaystyle \frac12\sum_{i=n}^k a_i\to\infty$ .

This is a contradiction, and indicates that repeated applications of this case always terminate, and lead to case 2. In particular, case 2 is considered infinitely often. Moreover, if $k>n-1$ is least such that $A_k-C_k>L$ , then $A_k-C_k=(A_{k-1}-C_{k-1})+(a_k-c_k)\le L+a_k$ .

If $A_{n-1}-C_{n-1}>L$ , then in particular $A_{n-1}\ne C_{n-1}$ , so at least one of the $a_i$ , $i<n-1$ , has not been considered yet as value of a $c_j.$ Pick the least such index $i$ , and define $c_n=a_i$ .

Since this case applies infinitely often, the value of the least index $i$ such that $A_i$ is not one of the $c_j$ increases unboundedly. By design, no index $i$ is used more than once, and therefore this algorithm results in a rearrangement of the $a_i$ . If we ever find an index $k$ such that (again) $A_k-C_k\le L$ , note that $A_k-C_k=(A_{k-1}-C_{k-1})+(a_k-c_k)>(A_{k-1}-C_{k-1})-c_k$ ${}>L-c_k$ .

Note that our assumption that the $a_k$ are decreasing ensures that, in case 2, (with notation as above) $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)\le$ $A_{n-1}-C_{n-1}$ , since $a_i\ge a_n$ . Moreover, it cannot be that we have equality from some point on, since the $a_k$ converge to $0$ . This means that the values $A_k-C_k$ decrease.

Since $a_i,c_i\to 0$ , it follows that, if case 1 is also applied infinitely often, then $A_i-C_i\to L$ . Therefore, to conclude, it suffices to argue that we cannot stay in case 2 forever. To see this, argue by contradiction, and assume that from $n-1$ on, we always stay in case 2. Let $j_m$ be the number of indices $i\le m$ such that $a_i$ is not one of the $c_k$ , $k\le m$ . For any $m\ge n$ , since $A_{m-1}-C_{m-1}>L$ , then $j_m\le j_{m-1}$ , with equality if and only if $a_m$ is not one of the $c_k$ , $k<m$ . Since $C_{n-1}\ne A_{n-1}$ , necessarily at least one of the $c_i$ , $i\le n-1$ , is $a_r$ for some $r\ge n.$ If $r_0$ is the least such index $r$ , then $j_{r_0}<j_{n-1}$ , because $a_{r_0}$ is one of the $c_i$ with $i<r_0$ , in fact $i<n$ . Since we cannot have an infinite decreasing sequence of positive integers, this means that (we have a contradiction and) eventually we should be in case 1 again. This completes the proof in the case the $a_i$ are decreasing.

Let’s now consider the general case. If we try to implement the argument we just described, we see that if we are ever in case 1, $A_k-C_k\le L$ , the sequence of $A_i-C_i$ increases until we find an index $j$ with $A_j-C_j>L$ . Also, once we enter case 2, we cannot stay there indefinitely, as the number of terms $a_i$ that are not some $c_k$ , $k\le j$ , decreases. The problem is that the sequence of $A_i-C_i$ is not necessarily decreasing. In fact, if we are at a stage where we define $c_n=a_i$ for some $i<n$ , and it happens that $a_n>a_i$ , then $A_n-C_n>A_{n-1}-C_{n-1}$ . This means that, although we have ensured that $\liminf (A_i-C_i)=L$ , and that there is a subsequence of the $A_i-C_i$ converging to $L$ , it may well be the case that $\limsup (A_i-C_i)>L$ .

Sierpiński deals with this situation in a clever fashion. He ensures that if we are in case 2, so that $c_n=a_i$ for some $i<n$ , then also $a_n=c_k$ for some $k<n$ , which means that $c_k$ was defined according to the prescription in case 1. He uses this to guarantee that $A_n-C_n$ cannot stray too far from $L$ . In detail, Sierpiński proceeds as follows, defining now three cases. Assume we are at the beginning of stage $n$ :

If $A_{n-1}-C_{n-1}\le L$ , as before let $c_n$ be some $a_r$ where the index $r$ has not yet been chosen, and $a_r<a_n/2$ , and $a_r<1/2^n.$
If $A_{n-1}-C_{n-1}>L$ , now we consider two possibilities, according to whether or not the index $n$ was picked previously: If it was not, then we let $c_n=a_n$ . Note that $A_n-C_n=A_{n-1}-C_{n-1}$ if this is the case, and that, since $C_{n-1}\ne A_{n-1}$ , then necessarily some $c_i$ with $i<n$ must be $a_r$ for some $r>n$ . This means that at some later stage (at most by stage $r$ ), we are no longer to be in this case.
If $A_{n-1}-C_{n-1}>L$ , and the index $n$ was chosen previously, then we let $c_n$ be $a_i$ , where $i$ is the first index less than $n$ not chosen yet.

As before, cases 1 and 3 happen infinitely often, so we have indeed defined a rearrangement. Moreover, if we are in case 3, then $a_n=c_i$ for some $i<n$ (this is why we included case 2). The point is that if stage $i$ is by case 2, then $c_i=a_i$ , and if it is by case 3, then $c_i=a_j$ for some $j<i$ . This means that the only way we can have $c_i=a_n$ for $n>i$ is if stage $i$ was by case 1, which means that $c_i<1/2^i$ .

Fix $\epsilon>0$ . Since cases 1 and 3 happen infinitely often, if we choose $n$ large enough, then we can ensure that all indices $m$ mentioned below are so large that $a_m,c_m<\epsilon$ , and if $a_m=c_k$ for some $k<m$ , then $k>N$ where $\sum_{i\ge N}1/2^i<\epsilon$ . We want to ensure that $|(A_n-C_n)-L|<2\epsilon$ . This proves that $A_n-C_n\to L$ .

If stage $n$ is by case 1, and stage $m<n$ was largest where $A_m-C_m$ ${}>L$ , then $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_{m+1}-c_{m+1})>$ $L-c_{m+1}>L-\epsilon$ . Since the sequence $A_i-C_i$ is increasing for $m+1\le i\le n$ , we have $L\ge A_n-C_n>L-\epsilon$ , as wanted.

Suppose then that stage $n$ is by case 3, and that stage $m<n$ was largest where we were in case 1. We then have that $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_m-c_m)\le L+a_m<L+\epsilon$ . Also,

$A_n-C_n=A_{m+1}-C_{m+1}+\sum_{i=m+2}^n(a_i-c_i),$

but for any such $i$ , we have that stage $i$ was either by case 2, and therefore $a_i-c_i=0$ , or else it was by case 3, and therefore $a_i=c_j<1/2^j$ for some $j<i$ , but necessarily $j>N$ , so

$\displaystyle \sum_{i=m+2}^n(a_i-c_i)=\sum\{a_i-c_i\mid m+2\le i\le n,$ and stage $i$ was by case 3 $\}$ $\displaystyle \le\sum\{a_i\mid m+2\le i\le n,$ and stage $i$ was by case 3 $\}$ $\displaystyle \le\sum_{j>N}\frac1{2^j}<\epsilon.$

This means that $A_n-C_n<L+2\epsilon$ . But also

$A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)>L-c_n>L-\epsilon.$

This completes the proof. $\Box$

This concludes the proof of Sierpiński’s theorem. $\Box$

Further work on rearrangements that fix pointwise prescribed subsets of $\mathbb N$ has been pursued by Wilczyński and others, partly in the context of descriptive set theory. See for instance

Rafał Filipów, and Piotr Szuca. Rearrangement of conditionally convergent series on a small set. J. Math. Anal. Appl., 362 (1), (2010), 64–71. MR2557668 (2010i:40001).

Parts of this post have been used in this MO question.

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3 Responses to 414/514 The theorems of Riemann and Sierpiński on rearrangement of series

Monica Agana – The classical theory of rearrangements | A kind of library says:

July 16, 2017 at 7:20 pm

[…] series of real numbers, prior results, and some extensions and related topics. This previous blog post discusses some of […]

Reply
hmeng1 says:

September 25, 2021 at 8:44 pm

Dear prof. Caicedo,

I just read your post on mathoverflow, and your post is closely related to my problem about “a restricted version of the Riemann series theorem: rearrangements with alternating signs” (I also use the same topic posted on mathoverflow), simply put, can we ask for an additional structure—the rearrangement is alternative, that makes the Riemann series theorem still hold?

I will appreciate it if you have time to help me, thanks in advance.

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ปั่นสล็อต says:

March 18, 2023 at 8:17 pm

ปั่นสล็อต

414/514 The theorems of Riemann and Sierpiński on rearrangement of series | A kind of library

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	What is a Cantor-sty… on 580 -Some choiceless results…
	Question about Gener… on 580 -Some choiceless results…
	Set has Measure Zero… on On strong measure zero se…
	Non-measurable subse… on 580 -Cardinal arithmetic …
	hmeng1 on 414/514 The theorems of Rieman…
	580 -Cardinal arithm… on 580 -Cardinal arithmetic …

A kind of library