mar1996p78-95_on the infinitude of the prime numbers
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GENERAL I ARTICLE
On the Infinitude of the Prime Numbers
Euler's Proof
Shailesh Shira1i
has been at theRishi ValleySchool
(Krisboamurti Foundation
oflDclia),Rishi Valley,
Andhra Pradesh, formore
than tea years and is
eurrendy the PrincipaL He
has been involved in the
Mathematical Olympiad
Programme since 1988.
He has a deep interest in
talking and writiDgabout
mathematics, particularly
about its historical
aspects. He is also
interested in problem
soIviDg(particularly in the
fields of elementary
number theory, geometry
and c:omblnatoric:a).
Adapted from the articles
InS4MASY..t Vol.2No.1.2.
78
Shailesh A Shirali
Euclid's elegant proof that there are infinitely many prime
numbers iswellknown. Euler proved the same result, in fact
a stronger one, by a1Ullyticalmethods. This Bfticlegivesan
exposition of Euler's proof introducing the necessary con-
cepts along the way.
Introduction
In this article, we present Euler's very beautiful proof that there
are infinitely many prime numbers. In an earlier era, Euclid had
proved this result in a simple yet elegant manner. His idea is easy
to describe.DenotingtheprimenumbersbyP1,P2,P3'... ,sothatp}= 2,P2 = 3,p3 = 5, on, he supposes that there are n primes in
all, the largest beingPII.He then considers the number N where
N =P}P2P3 onp,.+1,
and asks what the prime factors ofN could be. It is clear that N
is indivisible by each of the primes PI' P2'P3' on,P,. (indeed,
N:=I(modPi)for each i, 1:s;i :s;n). Sinceevery integer greater than
1 has a prime factorization, this forces into existence prime
numbers other than the Pi. Thus there can be no largest prime
number, and so the number of primes is infinite.
The underlying idea of Euler's proof is very different from that
of Euclid's proof. In essence, he proves that the sumof therecipro-
calsof theprimesisinfinite;that is,
III- + - + - + ... = 00.
PI P2 P3
RESONANCE~arch 1996
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GENERAL I ARTICLE
In technicallanguage,the series~j 1/Pjdiverges.Obviously,this
cannot possibly happen if there are only finitely many prime
numbers. The infinitude of the primes thus followsasa corollary.
Note that Euler's result is stronger than Euclid's.
Convergence and Divergence
A few words are necessary to explain the concepts of convergence
and divergence of infmite series. A series a} +az + a3 + . .. is
said to converge if the sequence of partial sums,
approaches some limiting value, say L; we write, in this case,
L i aj =L . If, instead, the sequence of partial sums grows with-
out any bound, we say that the series diverges,and we write, in
short, L i aj = 00.
Examples:
. The series1/1+ 1/2+ 1/4+ ... t: 1/2n + . .. converges (the
sum is 2, as is easily shown).
. The series 1/1 + 1/3 + 1/9 + . . . + 1/3n+ . . . converges (the
sum in this case is 3/2).
. The series1+ 1+1+ . .. diverges (rather trivially).
. The series 1-1 + 1-1+ 1-1+ 1- . .. also fails to converge, be-
cause the partial sums assume the values 1,0, 1,0, 1;0, ..., and
this sequence clearly does not possess a limit.
. A more interesting example: 1- 1/2+ 1/3 - 1/4 + ... ; a careful
analysis shows that it too isconvergent, the limiting sumbeingIn 2 (the natural logarithm of 2)
Divergence of the Harmonic Series .:E Iii
In order to prove Euler's result, namely, the divergence of ~ 1/p,.,
weneed to establish various subsidiary results. Along the waywe
RESONANCEI March 1996
Euler proved that the
sum of the reciprocals
of the primesis ,
infinite; the infinitude
of the primes thusfollows as a corollary.
A statement of the form I: 0,
= 00is to be regarded as mere-
ly a short form for the state-
ment thatthe sums 01' 0, + 02.
01'+02+' 03' do not possess
any limit. It is important to note
that 00 is not to be regarded asa number! We shall however
frequently use phrases of the
type',x = 00'Ifor various quanti-
ties xl during the course of this
article. The meaning should be
clear from the context.
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GENERAL I ARTICLE
The earliest
known proof of the
divergence of the
harmonic series is due
to the Frenchman
Nicolo Oresme and it
dates to about 1350.
80
shall meet other examples of divergent series. To start with, we
present the proof of the statement that
1 1 1 1-+-+-+ =00
1 2 3 4 .
This rather non-obvious result is usually referred to as thediver-
gena of the harmonicseries.The proof given below is due to the
Frenchman Nicolo Oresme and it dates to about 1350.We note
the following sequence ofequalities and inequalities:
1 1
2 =2'1 1 1 1 1-+->-+-=-3 4 4 4 2'
1 111 1 1 111-+-+-+->-+-+-+-=-5 6 7 8 8 8 8 8 2'
11 111 11
9 + 10+ ... + 16 > 16+ 16+ ... + 16 = 2'
and so on. This showsthat it ispossible to group consecutive sets
of terms of the series 1/1 + 1/2 + 1/3 + . . . in such a manner that
each group has a sum exceeding 1/2. Since the number of such
groups is infinite, it follows that the sum of the whole series is
itself infinite. (Note the crisp and decisivenatUreof the proof!)
Based on this proo~ wemake a more precise statement. Let S(n)
denote the sum
1 1 1 1
-+-+-+... +-1 2 3 n'
e.g.,S(3) = 11/6.Generalizing from the reasoning just used, wefind that
n
S(2") > 1 + 2.(3.1)
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GENERAL I ARTICLE
(Please fill in the details of the proof on your own.) This means
that by choosing n to be large enough, the value of S(2") can be
made to exceed any given bound. For instance, if we wanted the
sum to exceed 100, then.(3.1) assures us that a mere i98terms
would suffice! This suggests the extreme slowness of growth ofS(n) with n. Nevertheless it does grow without bound; loosely
stated, S( 00) = 00.
The result obtained above, (3.1),can alsobe written in the form,
Exercise:,!rite out a proof of the above inequality._ u _ _ _ __ _
A much more accurate statement can be made, but"it involves
calculus. We consider the curve.Q whose equation is y = l/x,
x >0. The area of the region enclosed by .0, the x-axis and the
ordinates x = 1 and x = n is equal to r ! dx, which simplifies1 x
to In n. Now let the region be divided into (n - 1)strips of unit
width by the lines x = 1, x = 2, x = 3, . . . , x = n (see Figure 1).
y
2x
3 4
RESONANCE I March 1996
The growth of5 (nl
with n is extremely
slow. Forthe sum to
exceed 100we
would require 2198
terms!
FlgUffI 1 7htI flgUffI Iht1w6
htJw to bound In n by 011-
MJI'VIng tht1t In n /$1hB IIIWI
IJI7dostJd by IhB t:UnIe Y =11%IhB x l1nd IhB tll'dl-
11III8 x = 1 IInd x = n.
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GENERAL I ARTICLE
In general, when
mathematicians find
that a seriesL a I
diverges, they are
also curious to know
how fast it diverges.
82
Consider the region enclosed by Q, the x-axis, and the lines x =i -I, x =i. The area of this region lies between 1/i and 1/(i-1),
because it can be enclosed between two rectangles of dimen-
sions 1x l/i and lx1/(i - 1), respectively. (A quick exAmination
of the graph will show why this is true.) By lettingi take the values2, 3, 4, . . . , n, and adding the inequalities thus obtained, we find
that
1 1 1 1 1-+-+... +-<lnn<-+-+2 3 n 1 2
1+ -. (3.2)n-l
Relation (3.2) implies that
1 1 1. 1 1 (3.3)lnn +-<-+-+-+...+-<lnn+l
n 1 2 3 n
and this means that we have an estimate for S(n) (namely, In n +
0.5) that differs from the actual valueby no more than 0.5.A still
deeper analysis shows that for large values of n, an excellent
approximation forS(n) isIn n + 0.577,but weshall not prove this
result here. It is instructive, however,to check the accuracyof this
estimate.WriteJ{n)forIn,n+ 0.577.Wenow find the following:
n=S(n)=
f(n)=
102.92897
2.87959
1000 100007.48547 . 9.78761
7.48476 9.78734
10000012.0902
12.0899
1005.18738
5.18217
The closeness of the values ofJ{n)and S(n) for large valuesofn is
striking. (The constant 0.577 is related to what is known as the
Euler-Mascheroni constant.)
In general, when mathematicians find that a series 1:ajdiverges,they are also curious to know how fast it diverges. That is, they
wish to find Ii function, say f (n), such that the ratio
(L 1"aj ) / f (n) tends to 1 as n ~ 00. For the harmonic series 1:1/i,
we see that one such function is given by f (n) = Inn.This isusuallyexpressedbysayingthat theharmonicseriesdivergeslike
the logarithmicfunction.Wenote in passingthat this is a very
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GENERAL I ARTICLE
slow rate of divergence, because In n diverges more slowly than
n£ for any £ > 0, no matter how smaU £ is, in the sense that
In n / n£ ~ 0 asn ~ 00for every£ > o. Obviously the function In
In n diverges still more slowly.
Exercise: Prove that if a > 1,then the series
converges. (The conclusion holds no matter how closea is to
1, but it does not hold for a = 1 or a < 1, a curious state of
affairs!) Further, use the methods of integral calculus (and ~
thefact that for a ::I;, the integral onh! is X(l-ll)(1 - a) toshow that the sum of the series lies between 1I(a-1) and
a/(a - 1).'< ~ =
The fact that the sum 1/1 + 1122 + "1132 + . .. is finitecanbe
shown in another manner that is both elegant and e1emen~:
We start with the inequalities, 22> 1x 2, 32> 2 x 3,42 > 3x 4,
. . " and deduce from these that
1 1 1 1 1 11 +-+-+-+...< 1 +-+-+-+...
22 32 42 1x 2 2 x 3 3 x 4
The sum on the right side can be written in the form,
1
(
! _ !) (! _ !
) (
! _ !)... (3.4)
+ 1 2 + 2 3 + 3 4 + ,
which (after a whole feast of cancellations) simplifies to 1 + 1/1,
that is, to 2.(This issometimes described bystating that the series
'telescopes' to 2.) Therefore the sum 1 + 1122+ 1132+ 1142+. . .is less than 2.We nowcallupon a theorem ofanalysiswhich states
that if the partial sums of any series form an increasing sequence
and areat the same time bounded, that is,they donot exceedsome
fixed number, then they possess a limit. We conclude, therefore,
that the series 1:1Ii2doespossess a finite sum which liesbetween
1and2.
RESONANCEI March 1996
The fact that the sum
1/12 + 1/22 +1/32 + ...
is finite can be shown
in a manner that is
both elegant and
elementary.
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GENERAL ARnCLE
The divergence of the
harmonic series
was independently
proved by
Johann Bernoulli in
1689 in a completely
different manner.
84
The divergence of the harmonic serieswas independendy proved
by Johann Bernoulli in 1689 in a completely different manner.
His proof is worthy of deep study, as it shows the counter-intui-
tive nature of infinity.
Bernoulli starts by assuming that the series 112+ 113+ 1/4+.. .
(note that he starts with 112rather 111)does have a finite sum,
which he calls S. He now proceeds to derive a contradiction in
the followingmanner. He rewrites eachterm occurring inS thus:
121113111
3"=6=6 + 6' "4= 12 = 12 + 12 + 12'. . . ,
and more generally,
1 n-l 1 1 1-= = + +...+-,n n(n-l) n(n-l) n(n-l) n(n-l)
with (n -1) fractions on the right side.Next hewrites the resulting
fractions in an array as shown below:
Note that the column sums are just the fractions .112,1/3,114,
115, . .. ; thus S is the sum of all the fractions occurring in the
array. Bernoulli now sums the rows using the telescoping tech-. -
nique used above (see equation (3.4». Assigning symbols to therow sums as shown below,
,1111111
A =2"+6+12+ 20 + 30 + 42 + 56 + ...,
1 1 1 1 1 1
B =6 + 12+ 20 + 30 + 42 + 56 + ...,
RESONANCE I March 1996
112 116 1112 1120 1130 1142 1156
116 1112 1120 1130 1142 1156
1112 1120 1130 1142 1156
1/20 1130 1142 1/56
1/30 1142 1156
1142 1156
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GENERAL I ARTICLE
1 1 1 1 1
C = 12 + 20 + 30 + 42 + 56 + " , ,
1 1 1 1
D = 20 + 30 + 42 + 56 + ...,
he finds that:
A = (1 - t) + (t - ~)+ (~ - i) + (~ - t) + .,.
= 1,
B = (t - ~)+ (~ - ~) + (i - t) + (t - ~) + .,.1
= -,2
1
C = 3' (arguing likewise),
1
D =4'
and soon. Thus the sumS,whichwehad written in the form
A + B + C + D + "', turns outto be equal to
Now.this looks disappointing -- just as thingswere beginningto
look promising!We seemto have just recovered the original series
aftera seriesofverycomplicatedsteps.Butin factsomethingsignifi-
cant has happened: an extra '1' has entered the series.At the startwe had defmedS to be 1(2+ 113+ 1/4+ "'; nowwe find that S
equals 1 + 112+ 113+ 114+ . ". This means that S =S + 1.
However, no finite number can satisfy such an equation. Conclu-
sion:S = oo!
RESONANCEI March 1996
Bernoulli's proof is
worthy of deep study,
as itshows the
counter-intuitive
natureof infinity,
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GENERAL I ARTICLE
The fundamental
theorem of arithmetic
states that every
positive integer greater
than one can be
expressed in precisely
one way as a product
of prime numbers.
86
There are many other proofs of this beautiful result, but I shall
leave youwith the pleasant task of coming up with them on your
own.Along the wayyou could set yourselfthe task ofproving ~at
each of the following sums diverge:
. 111+ 113+ 115+ 1/7+ 119+. . ';
. 111 + 1111 + 1121 + 1131 + 1141 + . . . ;
. 11a + lib + 1/c+ lid +. . .,wherea,b,c,d, ...,arethesuccessive
terms ofany increasing arithmetic progression ofpositive real
numbers.
Elementary Results
The next result that we shall need is the so-called fundamental
theorem of arithmetic: every positive integer greater than 1 can be
expressedin preciselyone way asa product ofprime numbers.We shall
not prove this very basic theorem of number theory. For a proof,
please refer to any ofthe well-known texts on number theory, e.g.,
the text by Hardy and Wright, or the one by Niven and Zucker-mann.
We shall also need the following rather elementary results: (i) if
k is any integer greater than 1, then
1 1 1 1 1- = 1 + -+ - + - + - +...1 - 14 k k2 k3 k4 '
which followsby summing the geometric serieson the right side,
and (ii) if aj , bj are any quantities, then
where, in the sum on the right, each pair of indices (i , J) occurs
precisely once.
Now consider the following two equalities, which are obtained
from (4.1) using the values k = 2,k = 3:
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GENERAL I ARTICLE
1 1 1 1 1- = 1 + -+ - + - + - + '"1- lJ2 2 22 23 .24 '
1 1 1 1 1-= 1 +-+ -+-+-+ ....1 - 113 3 32 33 34
We multiply together the corresponding sides of these two equa-
tions. On the left side we obtain 2 x 3/2 '=3.On the right side weobtain the product
(1 + 1/2 + 1122+ 1/23 + . . .) x (1 + 113 + 1132+ 1133+ . . .)
Expanding the product, weobtain:
1 1 1 1 1 11 + - + - + - + ... + - + - + - + '"2 4 8 3 9 27
1 1 1 1 1 1
+ (; + 12+ 24 + ... + 18+ 36 + 72 + ...,
that is, we obtain the sum of the reciprocals of all the positive
integers that have only 2 and 3 among their prime factors. The
fundamental theorem of arithmetic assures us that each such
integer occurs precisely oncein the sum on the right side. Thus we
obtain a nice.corollary: ifA denotes the set ofintegers of the form
]!l3b,where a and b are non-negative integers, then
. 1
L -; = 3...eA
ITwe multiply the left side of this relation by (1 + 1/5 + 1152+
1/53+ . . .) and the right side by 3;t1-1fs),weobtain the following
result:
1 3 ISL-=-=-'. z l-l/s 4Be B
. RESONANCE I March 1996
We obtain a nice
corollary: ifA denotes
the set of integers of
the form 2a3b,where
a and b are non-
negative integers,
then
L 1/z=3.ZEA
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GENERAL I ARTICLE
Euler was capable of
stunning
reasoning; some of
the steps in his proofs
are so daring that they
would leave today's
mathematicians
gasping for breath.
88
where B denotes the set of integers of the form l' 3bS', where a, b
andc denotenon-negativentegers.
Continuing this line of argument, we seethat infinitely many
such statementScan be made, for example:
. If C denotesthe setof positive integers of the form l' 3bSCI,
where a, b, c and d are non-negative integers, we then ~
have 1:zee 1/z= (IS/4) (7/6) = 3S/8.
. If D denotes the set of positive integers of the form
'f13b SCIlle, then 1:zeD 1/z= (3S/8)(l1/10) = 77/16.
Infinitude of the Primes
Suppose now that there are only finitely many primes, sayP1,P2,
P3' ...,P",wherePI = 2,p2= 3,P3= S,....We consider the product
111---1_1/21-1/31-1/5
This is obviously a finite number, being the product of finitely
many non-zero fractions. Now this product also equals
When weexpandout this product, wefind, bycontinuing the line
of argument developed above, that we obtain the sum of thereciprocals of aU the positive integers. To see why, we need to use the
fundamental theorem ofarithmetic and the assumption that 2,3,
S, .. .,p" areaUthe primes that exist; these two statementStogether
jmply that every positive integer can be expressed uniquelyas a
product ofnon-negative powersofthen primes 2,3,S, ...,p".From
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GENERAL I ARTICLE
this it follows that the expression on the right side isprecisely thesum
1 1 1 1
1+"2+3"+4"+ ...,
written in some permuted order. But by the Oresme-Bemoulli
theorem, the latter sum is infinite! Sowe have a contradiction:
the finite number
has been shown to be infinite -- an absurdity! The only way out
of this contradiction is to drop the assumption that there are only
finitely many prime numbers. Thus we reach the desired objec-
tive, namely, that ofproving that there are infinitely many primenumbers.
Note that, as a bonus, there are several formulas that drop out of
this analysis,more or lessascorollaries. For instance, wefind that
1 111
- ...=1+-+-+-+...1 - 1/52 22 32 42
that is, the infinite product and the infinite sum both converge
to the same (finite) value. By a stunning piece of reasoning,
including a few daring leaps that would leave today's mathema-
ticians gasping for breath, Euler showed that both sides of the
above equation are equal to 'fil/6.Likewise, we find that
1 111
-'"
= 1 + - + - + -+ ...,1- 1/54 24 34 44
and this time both sides converge to 1C4190.uler proved all this
and much much more; it is not for nothing that he is at times
referred to as analysisincarnate!
RESONANCE I March 1996
Itis not for nothing
that Euleris at times
referred to as analysis
incarnate!
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GENERAL I ARTICLE
Leonhard. ruler
Slnceunlvers~were not
the~f1lor researcb cen~
iresfJphlsdays, Leonhard
EdlerO~07-1783) spent
most of,hls.llfe with the
aerli~ and Petersburg
Academ.les..Plous,..bUtoot
dogmatic, Eulerconduct-
ed p,raxersfor his large
hou$abold, and creqfed
mathematiCswith.a.baby
oobls lap and cl1i1dreo
plt1Ylggall.around. Euler
,withheldhisownwotkon
calculu~ of "aria~o~ sothat young Lagrange
0736"'181~1ould publISh
ijflrSf, aodshowedslmllgr
generosityOIlmany other
occaslo!:,s.Utterly free of
fdlsepride," Euleralways
expla!OedhowhewaSled
to his results sayjog that
"the;ath 1 followed,.will
Perhapsbeofsomebelp".
And. 1ndE!ed,Qenera1lons
bfma1hei'na1ldansfollO'Ned
laplace's advice: "Read
Euier,he,1sour master In
aIU".
90
The Divergence of 1:lip
As mentioned earlier, Euler showed in addition that the sum
1 1 1 1 12, -=-+-+-+-+...P 2 3 5 7i ~ 1
is itself infinite. We are now in a position to obtain this beautiful
result. For anypositive integern ~2,let Pndenote the set ofprime
numbers less than or equal to n.Westart by showing that
n
1 1n ~ > L -:-.1- 4 JpeP. j=l
.
(6.1)
Our strategywill be a familiar one.Wewrite down the following
inequality for eachp EPn,which followsfrom equation (4.1):
1 1 1 1 1->1+-+-+-+...+-1 - 14 P p2 p3 pn.
The '>' sign holds becausewehave left out all the positive terms
that followthe term l/pn.Multiplying together the corresponding
sides 'ofall these inequalities (p EPn)'we obtain:
1
(
1 1 1 1
)n ->TI 1+-+-+-+...+-
1 - 14 P p2 p3 pnpeP peP. . .
When we expand out the product on the right side, we obtain a
sum of the form 1:j e A 1/j for some set of positive integersA. This
set certainly includes all the integers from 1 to n because the set
Pn containsall the prime numbersbetween1 and n. Inequality(6.1) thus followsimmediately.
Next, we already know (see equation (3.3))that
n
1 12,-:->lnn+->lnn.J n
j=l
(6.2)
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GENERAL I ARTICLE
Combining (6.1) and (6.2),weobtain the following inequality:
1n _ 1> Inn.
-~peP .
Taking logarithms on both sides, this translates into the state-
ment,
L In( 1~ 14)
> In Inn.
pe P ..(6.3)
Our task is nearly over. It only remains to relate the sum
~pePn1/pwith the sum on the left side of (6.3).We accomplish-
this by showing that the inequality
7x > In ~5 I-x
(6.4)
holds for 0 < x ~ 1/2.
To see why (6.4) is true, draw the graph of the curve r whose
equation isy = In (1/(1- x», over the domain - 00 < x< 1, (see
Figure 2). Note that r passes through the origin and is convex over
y
x
y=-In(l-x) x=l
RESONANCEI March 1996
~ Iv'" PrOdigiOUS;Output
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Ithas beer) estimated.1fKrt-;
Eole~s 886 works wpuld
fiU80 1a~gebooks. DldOt-
JogQrwritlog 00 hl~sl~,
Eulerkept lip his unporot.
IeledoUlputalUhrougnijis
life.~fjtofally~.~
thMost17yearsof~ ..
heprornlSeQJ9' c. tN.1.11he,,' ." ,.-~~
.~ AccKtenWwith,popers'!omUOyeors bJter
hisdeqlh; Qnecome ouf?9j
years;9fferhe diedf'~mostovAllll...~ .. ,;... ".:.:..J!I"""" '': '.,_", .,..,..: ",,-, . : .:..,~
rn~diedwhlle~.~il'.. LF-~l~~
,wiIh;bisgronddJildtef1 ~>. .'
drinlclng.. ~, "(All.boxed'
~onEu!er:1akenffom
'Gibe IBM poster Men of
Motl.~1ft;M'qthefflo!l~
1966.r"',f.\o; ,.-"L
Flgu1fI2 ",. gmph shows
that"" 0<.:;, <.:;,~ wehtwe
f21n 2Jx ~ In (II(1-xJJ.
91
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GENERAL I ARTICLE
The Genius of Eu'e,
Euler'swork on the .zefa
function, .partitions and
div~or sums remind us
thattle founded aoalyic
nUmber theory, w~lIe hiS
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GENERAL I ARTICLE
judge for themselves. LetS denote the sum ~j 1/Pj'We shall make
use of the following result:
eX ;:: 1 + x for all real values ofx,
with equality holding precisely when x =O.The graphs ofC and
1+x show why this is true; the fOrIilergraph is convex over its
entire extent (examine the second derivative of C to see why),
while the latter, a line, is tangent to the former at the point (0,1),
and liesentirely below it everywhere else. Substituting the values
x = 1/2, x = 1/3, x = 1/5, ..., successively. into this inequality, we
find that
Multiplying together the corresponding sides of these inequali-
ties, we obtain:
The infinite product on the right side yields the following series:
This series is the sum ofthe reciprocals ofall the positive integers
whose prime factors are all distinct; equivalently, the positive
integers that have no squared factors. These numbers are some-
times referred to as the quadratfreior square-freenumbers. Let Q
denote this sum. We shall show that this series itself diverges, in
otherwords,that Q = 00.This willimmediatelyimplythat S =00
(for eS> Q), and Euler's result will then follow.
We consider the product
(
1 1 1
)Qx 1+-+-+-+...
22 32 42
This product, when expanded out, gives the following series:
RESONANCE I March 1996
Johann Bernoulli
Johann Bernoulli 11667-
1748), tI'Ii:!youngerbrother
of Jal<ob Bernoulli11654-
1705) took it upon himself
to spread Lelbnlz's calcu-
lus across the Europeancontln.ent. Johann's proof
of the divergence oftl'le
harmonic series ~rst ap-
peared11689) In Jakob's
treatlse"anclwlth unchar"
acteristlc fraternal affec-
tion, Jakob even prefaced
the.argumel'lt with, an
acknowledgement of his
brothe(s priority.
93
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GENERAL I ARTICLE
!II ~
GHIIardr;JiM,Wqpt.,qIntroduction to the
n~ ofNumber;4th.n-&-a~
~~"Ivan Niven. Herbert S
Z1IekermamLAa:fatro-...
cJuctioa to the .~ofNUmben..wne,Eut-
~emI':!td..1989.'
TOIDApostoLAa:~
tiOli.to AaaIytic ~bet" Theol'J. Narosa
W Boase.I9't9.
,94
1 1 1 1-+-+-+-+...1 234 '
that is, we obtain the harmonic series. To see why, note that every
positive integer n can be uniquely written as a product of a square-free number and a square; for example, 1000 = 10 X 102,2000 =
5 x 202,1728 = 3 x 242,and so on. Now when we multiply
(
1111111111
)+2+3+5+6+7+ 10+11+13"+ 14+15+...
with
(
1 +1-+1-+1-+ ...
J2 32 42
we find, by virtue of the remark just made, that the reciprocal of
each positive integer n occurs preciselyonce in the expanded
product. This explains why the product is just the harmonicserieS.Now recall that the sum
is finite (indeed, we have shown that it is less than 2). It followsthat
Q x (some finite number) = 00.
Therefore Q = 00, and Euler's result (1:;l/p; = 00) follows. QED!
Readers who are unhappy with this style of presentation, in
which 00 is treated as an ordinary real number, will find it an
interesting (but routine) exercise to rewrite the proof to accord
withmore exacting standards of rigour and precision.
Conclusion
Amuch deeper - but alsomore difficult - analysis showsthat thesum 1/PI+ 1/P2 + 1/P3 + ... + I/P,. is approximately equal to
RESONANCEMarch 1996
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GENERAL I ARTICLE
x VL S.",m. {nit; inftnit. b.rmo,,;upr.,.,lff J t + f +1+ I + ~ &&. ~R ;"G..i...J 4 $ "'JH-
Id primus deprmcndir Frater :- invenranamque per pr2Ced~
(umma C~ieif :+-fr+ 1\+ 1~ + t5 J &e. viCurusporro, quid emer-seret ex JRa Cene J f + i + 1\+ ::~+ io ' &e. fi reColvel'CtUl'me-thodo Prop. X 1v. collegit p opofitionJS veriratem ex abfurditatcmanifefia, quae fequetetUr, ti Cumma.Cetia. harmonicae 6nira fiawe-.mur. Animadvenit: eniin II
Setiem A, t + t+ i + f + t+ t. &~, 3) (fi.ttionibus 6nguli':
iD alias J quarum numeratores funt I, Z JJ, 4-, &~ transmuratis)
iCrieiB,f+.r+I~ + :.t+-Io+4~ J&ce.X'C+D+E+F.&t..- -c.j+l+I~+1\+~7~J&c-..:x> perprac.il'D...+H-n+~~+T~+4~'&~ :X>C'-f:X>f
JE. . . +T!+Y5+~+4\ J &~:x>D- ~:x>t ~d~;'F. . . . .. "+"ik+1k+;p;~&ce.:X>E-TI:x>4. fequi..
&e.3'\ &c. wr,te..
(riem G 3) .of~ totUm patti J fi flUlUlll finita. elf~
Ego,
In In n. This is usually stated in the following form: as n tends
to 00, the fraction
14 +14 +14 +...+14. 2 3 II
lnlnn
tends to 1. This is indeed a striking result, reminiscent of the
earlier result that 111 + 112 + 113 + . . . + IIn is approximately
equal to In n. It shows the staggeringly slowrate of divergence of
the sum ofthe reciprocals ofthe primes. The harmonic series 'r.jlli
diverges slowly enough - to achieve a sum of over 100, for
instance, we would need to add more than 1043terms, so it iscertainly not a job that one can leaveto finish offover a weekend.
(Do you see where the number 1043comes from?) On the other
hand, to achieve a sum ofover 100with the series 'r.jIIpj,weneed43
to add something like 1010 terms!! This number is so stupen-
dously large that it is a hopeless task to make anyvisual image of
it. Certainly there is no magnitude even remotely comparable toit in the whole of the known universe.
JDhtlnn'$d/wNgent:flpn»f,""", JllkDb'$Tractatus de
serfebus Inflnltls, rsPIlb-
//shed /n ma. (FnHn pt1I/fI
'" t1f Journey through Ge-nius by Will/11mDllnhtlm J.
Address for correspondence
Shailesh ShiraI!
RishlValley School
Rlshl Valley 517 352.
Chlltoor Dist. IA PI. India.
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