Rearranging Terms in Alternating Series
Author(s): Richard Beigel
Source: Mathematics Magazine, Vol. 54, No. 5 (Nov., 1981), pp. 244-246
Published by: Mathematical Association of America
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Rearranging
Termsin Alternating
Series
RICHARD BEIGEL
StanfordUniversity
CA 94305
Stanford,
It is wellknownthatifyouchangetheorderof thetermsin a conditionally
convergent
series,
thesumof a
theresultmayhavea different
sum.We willdemonstrate
a methodforcalculating
broadclassofsuchreordered
series.The resultis notnew;itoriginally
appearedin a 19thcentury
Germanpaper[8],and similarworkappearedlaterin [2],[4],and [6].The purposeof thisnoteis
to bringattention
to theresult,
whichcouldappropriately
be mentioned
in introductory
calculus
classesrightafterRiemann'stheorem
morecomplicated
thanthe
[1]; themainresultusesnothing
integral
test,and thediscussion
usesnothing
morecomplicated
thantheconceptof o(f), which
couldbe explained(see [5]) or circumnavigated
in lecture.
The resultsfrom[3] on thealternating
harmonic
seriescouldwellbe includedin thelecture.
Let S = I' (-l)i+'aj, witha1 > 0 and {a1} eventually
to zero.
monotonically
decreasing
DefineSm' as therearrangement
(no signschanged)obtainedfromS by takinggroupsof m
positivetermsfollowed
terms(thusS" S and S21 a, + a3-a2 + a5
bygroupsof n negative
+a7-a4
+ *--).
THEOREM. Letf be a continuous
and eventually
monotone
suchthatf(2k - 1)
function
forpositiveintegral
k, andletm > n> 0. Then
Sm = S +
a2k
I
lim i f(x) dx,
2 k-oo0,k
ormaydiverge
withtheunderstanding
thatbothsidesmaybe infinite
byoscillation.
Proof.Sinceak
-
0, Stn is thelimitofits(m + n)kthpartialsum.Identically,
rnk
S(m+n)k = S2nk +
2
j=nk+ I
(1)
a2j1.
Sincef is eventually
monotone
to zero,thelimit
decreasing
t
lim : f(2jf-1)- f(2x-1)dx
It followsthatthe
testforconvergence.
exists;thisfollowsfromthestandard
proofoftheintegral
tailexpression
goesto 0; therefore,
lim j
k-??
/ rk
a
j=nk+?I
rnk
a2j-
I
Pk
f(2x-
1)dx
0.
Therefore,
lim
k-?
244
lika2
V= nk+ I
kk-oo ,k
f(2
-
1) dx
MATHEMATICSMAGAZINE
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I
I
f(x)dx
lik
mk-f
k-oo 2 2 tk- I
=
kim fkf(x)
2 k-&oo sk
(2)
dx,
Let
or maydivergeby oscillation.
thatbothlimitsmaybe infinite
againwiththeunderstanding
k - oc in (1) and use (2) to finishtheproof.
"= S. If
If (aj) is absolutely
convergent,
thenwe get theexpectedresultthatSn
REMARK.
rearrangem= n we also find,as expected,thatS""' = S. (Severalless obvioussum-preserving
mentsofinfinite
seriesarediscussedin [7].)
the diverseresultsthatcan be obtainedby applyingour
Severalexampleswill illustrate
Sm" ofparticular
series.
alternating
Theoremto rearrangements
+
EXAMPLE1. If H""' is obtained fromthe alternatingharmonicseries (1 - 1/2 + 1/3 - 1/4
) by takinggroupsof m positivetermsand n negativeterms,then
1.
lim|
H 'n=log2+
J-dx =log2+jlog-2log
kl
I
4m
seriesI - 1/2P + 1/3P- 1/4P
fromthealternating
EXAMPLE
2. If pnn is obtainedsimilarly
"'
to positiveinfinity.
+
with0<Op < 1 and m> n, thenP diverges
. Then
EXAMPLE3. Let L be Leibniz's sum for</4: L 1 - 1/3 + 1/5 - 1/7 +
.**,
7
=
+ 2I lm
Ltt Mi
4
2 k-J0
EXAMPLE4. Let S - 1,= (-l)i+
St" =S?+
EXAMPLE5. Let S=
thatf is monotone.
S ?+
smn
irkI
(1/(2x-l))
/k
k
410g-.
I dx='4+
4 4
nl
'/jlog j. Then
lim (loglogmk-log lognk) = S.
2 k-oo
wheref(x) =(cos(logx)
,=,(-l)j+lf(j),
+ 2)/x. It is easy to check
lim (sin(log mk) + 2log mk-sin(lognk) -2log nk),
k-oo
unlessm= n.
whichdiverges
byoscillation
to a
ifwe can findan exampleforwhichS"n"converges
Thecuriousreadermaybe wondering
functions
"artificial"
term.Barring
otherthana logarithmic
fromS by something
valuediffering
thereis none.
forwhichlim._Oxf(x) diverges
byoscillation,
Proof.Iff(x) = o(l/x), then
lim
f( x) dx 6 klim (imk-nk
-oo
k -00 fk,k
)f(
nk),sincef is monotonedecreasing,
m
nkf(nlk),
fl
=0.
Therefore
Smn= S. Iff(x) = c/x + o(l/x), then
limkf f(x)
k-00 tk
dx - lim
k-00
,k
= lim |
?+ (-) dx
X
k-00 'nk X
= clog
X
dx, by theprecedingresult,
n,
n
1981
VOL. 54, NO. 5, NOVEMBER
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245~~~~~~~~~~
if
so S" =S++Lclog(m/n). The readermay check that S"m' divergesto positiveinfinity
>
f(x) O(l/x).
The remaining
possibility,
thatxf(x) diverges
byoscillation,
seemsto meritattention.
Suppose
we modifyExample5 slightly
by letting
f(x) = (cos(alogx) + 2)/x, wherea 2-7/log(m/n).
Then
alog mk- alog nk= alog
2 T, so sin(alogmk) - sin(alognk) =0.
-=
n
Hence
S'l = S +
I
+
lim
mk- sin(alognk) - 2lognk) = S + log-.
2 k-. (sin(alog mk) 2log
fl
is proportional
ifand only
Onceagainthedifference
tolog(m/n).Also,noticethatSpq converges
ifalog( p/q) is an integral
ifand onlyifp/q is an integral
multiple
of21r,or equivalently
power
ofm/n.It wouldbe interesting
to findanswersto thefollowing
twoquestions:
S" -S converges
Is there
a seriesS forwhich
forsomevaluesofmandn,butis notproportional
tolog(im/n)?
Is therea seriesS= ,=(-l)i+'f(j)
such thatxf(x) divergesby oscillation
butSt" -S
converges
forall valuesofm andn?
EXAMPLE
6. SupposeS = 2 1(- l)'+sin(I/j). Naturally
we takef(x) to be sin(l/x),which
is I /x + o(l/x). Thuswe findthat
= S+ I log- lim xsin-= S + I log-.
Sm11
to
backtoourtheorem,
thatm> n is unnecessarywe shouldnotethattheassumption
Getting
to
see this,just identify
In
a
modification
with
a21fact,
slight
(left
j=mk+
X=}nk+1a2j-1
thereader)ofourproofyieldsa similarresultforanyrearrangement
ofS thatleavestheorderof
its positiveand negativesubsequences
of S withf( k)
intact;if S' is such a rearrangement
ourold hypothesis,
termsin Sk, andf satisfies
then
negative
+I
1
li
1k-oo
-0k-f (x
dx)
(k)
thisresultto theseriesS = 1-2 - 1/2 + 3 -1/2 - 4 - 1/2 +
*,and taking
Applying
'p(k) tobe
the integralpart of ((I- 2k - 12 )/2)2 fork > 112 (p can be arbitrary
a simple
elsewhere),
showsthattherearranged
calculation
seriessumsto S + 1.
For anotherapplication
of thisresult,
choosethereordering,
H', of thealternating
harmonic
seriesforwhich?(k) equalstheintegral
partof k/(4e2 + 1). ThenH' = 7T.
I wish to expressmy thanksto all thosewho read over and commentedon thismanuscript,
especiallyAlan
Siegel.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
246
T. Apostol,Calculus,vol. I, 2nd ed., Blaisdell,MA, 1967,p. 413.
T. J.I'A. Bromwich,
An Introduction
to theTheoryof InfiniteSeries,2nded.. Macmillan,London,1947,pp.
74-76.
C. C. Cowen,K. R. Davidson,and R. P. Kaufman,Rearranging
thealternating
harmonicseries,Amer.Math.
Monthly,87 (1980) 817-819.
K. Knopp,Theorieund Anwendungder unendlichen
Reihen,5thed., Springer-Verlag,
Berlin,1964,p. 335
(problem148).
J.Olmsted,AdvancedCalculus,Appleton-Century-Crofts,
NY, 1961,p. 141.
A. Pringsheim,
Ueber die Werthveranderungen
Reihen und Producte,Math. Ann.,
bedingtconvergenter
22(1883)455- 503.
Paul Schaefer,Sum-preserving
of infinite
rearrangements
series,Amer.Math. Monthly,88(1981) 33-40.
0. Schl6milch,
Ueberbedingt-convergirende
Reihen,Z. Math.Phys.,18(1873)520-522.
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