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



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.

79



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)

RESONANCE I March 1996



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


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.

81



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




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.


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.

83



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 + ...,


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

1156



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!


Bernoulli's proof is

worthy of deep study,

as itshows the

counter-intuitive

natureof infinity,

85



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:




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

flI



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




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!


Itis not for nothing

that Euleris at times

referred to as analysis

incarnate!

89



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)




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


~ Iv'" PrOdigiOUS;Output

. 'it

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



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



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:


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



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



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.

Documents

Mar1996p78-95_On the Infinitude of the Prime Numbers