A Note on λ-Permutations Author(s): Daniel J. Velleman Source: The American Mathematical Monthly, Vol. 113, No. 2 (Feb., 2006), pp. 173-178 Published by: Mathematical Association of America Stable URL: http://www.jstor.org/stable/27641871 Accessed: 25/01/2010 13:59 Your use of the JSTOR archive indicates your acceptance of JSTOR's Terms and Conditions of Use, available at http://www.jstor.org/page/info/about/policies/terms.jsp. JSTOR's Terms and Conditions of Use provides, in part, that unless you have obtained prior permission, you may not download an entire issue of a journal or multiple copies of articles, and you may use content in the JSTOR archive only for your personal, non-commercial use. Please contact the publisher regarding any further use of this work. Publisher contact information may be obtained at http://www.jstor.org/action/showPublisher?publisherCode=maa. Each copy of any part of a JSTOR transmission must contain the same copyright notice that appears on the screen or printed page of such transmission. JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms of scholarship. For more information about JSTOR, please contact [email protected].

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gis)

=

cos s ? cosias)

then gw(0) = g"iO) = g'(0) = giO) = 0, while

cosibs),

g""i
= -4a2b2.

? ? it follows that there exists S > 0 such that Taking a 1/2 and b V3/2, < cosiv^?), 0 = g(0) > g(2?) when 0 < 0 < 8. This gives cos0 = cos(2?)/cos(9 whence 0 > \/3# as claimed. of these arguments (including the Incidentally, we remark that simple modifications use of the hyperbolic cosine rather than the cosine in the calculus lemma) yield a proof that the hyperbolic plane is not (locally) flat. Department

of Mathematics

University

of Florida FL 32611-8105

Gainesville, robinson @ math. ufl. edu

A Note

on ?-Permutations Daniel

In their paper [3], Steven Krantz ations. They define a permutation two properties: has the following

J. Velleman

and Jeffery McNeal study what they call X-permut a of the positive integers to be a ?-permutation if it

1. For every convergent series Y1T=\ ak>trie series Y1T=\ ae(k) also converges. 2. There is at least one divergent series ^at Y1T=\ a^ suc^ 1lT=\ a^(?) converges. We call a permutation a with property 1 a convergence-preserving permutation. Notice that property 2 is equivalent to the statement that a-1 is not convergence-preserving. if and only if it is convergence-preserving, Thus, a permutation a is a ?-permutation but its inverse is not. We begin with a characterization of convergence-preserving The permutations. characterization we give can be found in [5], where it is derived by combining results of R. P. Agnew is more direct. A similar [1] and P. Vermes [7], but our derivation can also be found in [4], and related results can be found in [2] and characterization [6]. We will need the following notation: if c and d are positive integers and c < d, then [c, d]z =

{x e Z+

: c < x < d}.

Suppose now that a is a permutation of the positive integers. For any positive integer n we can write {
a (2),

..., a in)} =

[cnx,d?]z U [cn2,dn2}z U

U

[cnbn,dnbJz,

(1)

> where c" and d" (1 < / < bn) are positive integers, c" < ?/f, and 2 when c"+1 d? + ? 1 and ever i < bn. Notice as that, since a is onto, we can make c\ large as we d\

February

2006]

NOTES

173

please by choosing

n sufficiently =

Mn

+ dnbn

large. Let 1 = max{or(l),

or(2),

..., a(n)}

+

1.

say that bn, the number of blocks in the union in (1), is the nth block number for a and that {bn}%L{ is the block number sequence for a. Finally, we call a sum-preserving tne sei"ies if for every convergent series and the Y1T=\ a^ YlkLi a?(k) a^so converges, two series have the same sum. With this terminology in place, we can now state our

We

characterization

Theorem

(a)

a

convergence-preserving

(c)

a permutation

concerning

a of the positive

in

equivalent: is convergence-preserving.

(b) The block number a

permutations.

statements

1. The following

are

tegers

of

sequence for a

is bounded.

is sum-preserving.

Proof. Clearly (c) =>> (a); we prove that (a) =^ (b) and (b) => (c). (a) => (b): Suppose that the block number sequence for a is unbounded. We will find a convergent series Y1T=\ ak sucn tnat YlkL\ a?(k) diverges. 1 < k < Mni as fol Choose rt\ such that c\x ? 1. Now define ak for k satisfying lows: 1 ak ?

Notice

\?\ 0

if k ? d"1 for some / with 1 < / < bn{, if k = d"1 + 1 for some / with 1 < / < bn], otherwise.

that

Y^acr(k) k=\

=

> bnx 1,

YLUk k=\

= 0'

while Yll=\ ak is either 0 or 1when 1 < n < Mn{. Proceeding recursively, we will now choose for each j > 1 a positive integer n? such that Mn. > Mn._x, and we will also define ak for k satisfying Mn._x < k < Mn.. At each step, we will make our choices so as to ensure that Mn:

Y^ak=0 k=\ Suppose that j > 1, that a positive integer n}_\ has been chosen, been chosen for k satisfying 1 < k < Mn._x in such a way that

and that ak has

Mn._x

ak= 0. ^2 ?:=l Since Mn._j

174

the block number sequence for a is unbounded, we can choose n? so that dx and bn. > j2. We define ak for k satisfying Mn ._1< k < Mn. as follows:

?

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ak

=

ifk = d"J for some / with 1 < / < bn., some / with 1 < i < bn., if k = d\J + 1 for otherwise

1/j ? 1/j 0

Then j

=?

Mflj

bn.

^ao{k)

> ;,

7

?:= 1

^fljk

= 0,

?=1

and XX=i ?it is either 0 or 1/j when Mn. _1 < n < Mn.. ? This completes the definition of the series 0, but Clearly J2T=\ ak E?lia?ES=i a?(k) diverges. (b) => (c): Suppose that the block number sequence for a is bounded and that ? ? L. n letS(n) = ?jfe.Thenlim?^00 Sin) Y^Li ctk L. For each positive integer E*=i For any n sufficiently large that c\ ? 1,we have by equation (1) u

dn

?aff(i) it=l

dn

= + ?flJt ? /=2?flt fc=l

yt^J2

u

= /=2

5K)+?(swr)

s(c" !))

-? oo as ^ -> oo, so all of the values of S in the last sum in d\ equation approach L. Since bn is bounded, it follows that

Now

= Eaa(k) as

lim

J n->oo y^^a(^

(2) (2)

= L,

required.

Theorem and McNeal

1 resolves

two issues from [3]. The first is that on page 33 Krantz that it would be of interest to investigate the sum-preserving suggest as a subset of to Theorem all However, ?-permutations ?-permutations. according are sum all ?-permutations 1, this subset is in fact the entire set of ?-permutations: a it is evident from if that is Second, preserving. equation (2) convergence-preserving, then there is a positive integer B such that, when the terms of a series are permuted by a, every partial sum of the permuted series can be written as a linear combination with coefficients ?1 of at most B partial sums of the original series. In other words, letting Sin) denote the nth partial sum of the original series, for each n there is a < B, a positive integer j satisfying j sequence of positive integers m\, m2, ..., m;-, and a sequence of coefficients c\, c2, ..., cj; in {? 1, 1} such that n

^2a?(k) k=\

j

=

^CiSimi). ?=1

We can take B to be twice the bound on the block number sequence for a. This shows that the example in the second paragraph of section 5 of [3] is incorrect. In section 3 of [3], Krantz and McNeal show that there are uncountably many but they do not compute the exact cardinality of the set of ? ?-permutations, In fact, this is not hard to do. As in [3], we let J\f be the set of all permutations. a such that both a and a~x are and ? the set of all permutations ?-permutations, convergence-preserving.

February

2006]

For each set X of positive NOTES

integers

let rx be the permutation

175

? 1 and ? of the positive integers that swaps 2m 1 and 2m if m is in X and leaves 2m It is easy to see that for every n the nth block number for rx is 2m fixed otherwise. It is also clear that xx is its either 1 or 2, and therefore xx is convergence-preserving. own inverse, so xx belongs to?. Furthermore, the permutations rx for different X are different, so this is a family of 2K? distinct elements of ?. Thus, card(?) > 2^?. But the set of all permutations of the positive integers has cardinality 2K?, so it follows that = 2*?. < 2^? as well, and therefore card(O) card(O) let a be any ?-permutation. of two It is easy to see that the composition Now convergence-preserving

gence-preserving

then

convergence-preserving,

so xx

is convergence-preserving,

permutations

for every set X of positive o rx

a~x

integers.

o xx =

a~l

would

o a

is conver

If (xx o a)"1 = o~~l o xx were be

as well,

the

contradicting

fact that a is a ?-permutation. Therefore rx o a is amember of ?? for every X. Clearly these are all distinct, so they are 2^? distinct elements of ??. As before, it follows that = 2K?. Thus, we have established theorem: the following card(A0 Theorem

2. card(AT) =

=

card(O)

2^.

Finally, we give examples of two infinite series whose behavior with respect to is interesting and state an open problem motivated by these examples. ?-permutations but the The terms in both of these series are the numbers ?1, ?1/2, ?1/3,..., terms are arranged in different orders in the two examples. In both series, the pos itive terms appear in the order 1, 1/2, 1/3,... and the negative terms in the order ? the only difference between the series is in how the positive and 1, ?1/2, ?1/3,...; negative terms are interleaved. For the first example, we alternate blocks of positive terms with individual negative terms, and we make sure that each block of positive terms adds up to at least 2. Thus, the series begins

like this:

fi

11 33

11 56

111 234

?

2

sequence of partial sums for this series goes up by at least two for each block of positive terms and down by at most one for each negative term. It is clear, therefore, that the partial sums approach infinity, and the series diverges. Of course, the terms can be rearranged tomake the series converge; we might say that it is a "conditionally

The

divergent

series."

Now, consider the result of permuting the terms of this series by some ?-permutation a. Let B be a bound on the block number sequence for a. It is not hard to see that ? 1. the sum of a consecutive sequence of terms in this series cannot be smaller than Therefore, as in equation (2), we obtain for sufficiently large n bn

d'{

^aa{k) k=\

=

^ak jfe=l

+

d?d?

df

Y^Ylak i=2

k=c^

-

~ Ylak k=\ k=l

~ (bn ^ - XIa*

~~

(B

-

!)

?> oo as n -> oo, the partial sums of the permuted series also approach dnx the This shows that, although infinity. original series can be made to converge by per cannot the be this done terms, muting by using a ?-permutation. For our second example, we alternate blocks of positive terms with blocks of nega tive terms. We choose these blocks so that the first block consists of positive terms that add up to s\, where s? > 1, the next block consists of negative terms adding up to ?1{, Because

176

?

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t\ > s\ + so on. and Thus,

where

? ak > /c=l

1, the next block is positive the series starts like this:

=

1

-

f?

11111 2 3

1-1-1-1-1-.

terms adding up to s2, where

11 33

423

s2 >

t\ +

1,

5

? The partial sums of this series first rise to s\ > 1, then fall to s\ t\ < ?1, then rise to ? ? > < so on. Clearly this 1 to then fall and + + + Si s2 si t\ s\ s2 >2, t\ t2 ?2, is another conditionally series. divergent a such that, Now suppose that L is any real number. We will find a ?-permutation when the terms of the series are permuted according to a, the sum of the resulting series is L. We begin by finding a number m \ such that ami > 0 and m\

L<

y^aic

< L + 1.

k=\

Such an m\ must exist because of the way the partial sums swing back and forth between larger and larger magnitude positive and negative numbers and the fact that at most 1. The first m\ terms of the permuted all terms in the series have magnitude ? k series are the same as those of the original series. In other words, we let o{k) sum For is when k
in that proof,

it is clear that the resulting

n L -

1<

^aa(k)

< L+

1

k=\

if n > mx. What

remains

to be shown

is that the permutation

o generated

in this way

is a ?-permutation.

Notice that in the permuted series the first term of the j th negative block is used at step m\ + 1,which occurs before the step at which the first term of the (7 + l)th pos itive block is used. Also, the first m\ terms of the permuted series include all terms of the original series before the 7 th positive block, as well as some terms of this positive block. We claim now that in the permuted series the first term of the (7 + l)th positive block is used before the first term of the (7 + l)th negative block. To see why, suppose not. Then the first term of the (7 + l)th negative block is used first, say as term number mr + 1 of the permuted series. This means that for k satisfying mx < k < m' term number k of the permuted series must come from either the 7 th positive block or the 7 th negative block and that these terms must include all terms in the 7 th negative since term is constructed, block. Also, according to the rule by which the permutation sum number m' + 1 is negative, the of the first m' terms of the permuted series must

February

2006]

NOTES

177

be greater

we have

than L. Consequently, m

L <

m\

< ^aa[k)

+

^2aa(k)

k=\

s,

~ tj

< L+

1-

1= L,

k=\

is a contradiction. Thus, the first term of the (j + l)th positive block is used at some step m2 + 1 that comes before the step at which the first term of the (j + l)th negative block is used. the first m2 terms of the permuted series include all terms of the original Moreover, series before the y th negative block, as well as some terms of this negative block. Similar reasoning can be used to show that the first term of the (j + l)th negative which

the first term of the (j + 2)th positive block. Indeed, by induction that the first term of every block is used before the first term of the next block. Since both the positive terms and the negative terms are used in order, it is not hard to see that the block number sequence for the permutation a is bounded 1, a is a ?-permutation. by 2. Thus, by Theorem

block is used before we can demonstrate

These examples suggest the following question, to which we do not know the an swer. Suppose that *s a conditionally divergent series. We know that this series Y1T=\ ak can be rearranged to converge to any real number. But what if we restrict our attention to ?-permutations? Let

S=

\L e R

: for some ?-permutation

a,

[ We have seen an example anything in between?

in which

S

?

0, and another

V]

ao{k)

= L \ .

k=\

in which

J S = R. Can S ever be

REFERENCES 1. R. P. Agnew, On rearrangements of series, Proc. Amer. Math. Soc. 6 (1955) 563-564. U. Elias, Rearrangement of a conditionally 110 (2003) 57. series, this Monthly convergent 3. S. G. Krantz and J. D. McNeal, 111 (2004) 32-38. series, this MONTHLY Creating more convergent 4. R A. B. Pleasants, Rearrangements that preserve convergence, J. London Math. Soc. 15 (1977) 134-142. 2.

5. 6.

P. Schaefer, Sum-preserving rearrangements J. Stef?nsson, Forward shifts and backward this Monthly

7.

P. Vermes,

111

(2004) Series-to-series

of infinite shifts

88 (1981) 33-40. series, this Monthly in a rearrangement of a conditionally convergent

series,

913-914. transformations

and analytic

continuation

by matrix

methods,

Amer. J. Math.

71(1949)541-562.

Department of Mathematics Amherst College Amherst, MA 01002 djv ell eman@ amherst.

178

?

and Computer

Science

edu

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