DIRICHLET'S TH50RSM ON PRIMES IN AN

ARITHMETIC PROGRESSION

by

THELM ANI ' GHENAULT, B.S.

A THESIS

IN

MATHEMATICS

Submitted to the Graduate Faculty of Texas Technological College in Partial Fulflllnent of

the RequireTisnts for the Degree of

MASTER OF SCIENCE

Approved

Accepted

May, 1966

PK'CG- \\\\J^

A/o. \2

ACKNOWLEDGMENT

I am deeply indebted to Dr. Anthony Gioia for his

direction of this thesis.

ii

TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS ii

INTRODUCTION 1

PART I. SPECIAL CASES OF DIRICHLET'S THEOREM

Chapter

I. ELEMENTARY RESULTS 3

EUCLID'S PROOF

PROOF OF THE GEOMETRIC PROGRESSION ^n+3

PROOF OF THE GEOMETRIC PROGRESSION 6n+l .

II. ANALYTIC RESULTS 7

EULER»S PROOF

DIRICHLET'S PROOF OF THE PROGRESSION kn+1

Part II. THEOREMS NECESSARY TO THE PROOF OF

DIRICHLET»S THEOREM 21

III. CHARACTERS AND THEIR PROPERTIES 22

IV. THE GENERAL THEORY OF DIRICHLET SERIES 30

Part III. DIRICHLET'S THEOREM AND SOME OF ITS

CONSEQUENCES 51

V. PROOF OF DIRICHLET'S THEOREM 52

VI. CONSEQUENCES OF DIRICHLET'S THEOREM 66

LIST OF REFERENCES 70

iii

INTRODUCTION

This paper contains an extensive and detailed proof

of Dirichlet's theorem via analytic methods. This famous

theorem states that in any arithmetic progression kn+a,

n=0, 1, 2, .,., where k, a, are positive integers, having

no common factors gre -ter than one, there is an infinity

of primes.

Some special cases of Dirichlet's theorem were proven

much earlier by Euclid, Suler, and others. Euclid proved

that there are an infinite number of primes of the form of

2n+l by elementary methods, and two thousand years later

Euler gave an analytic proof of this same theorem.

Dirichlet (I805-I859) patterned his method of proof

after that of Euler. Dirichlet's proof by analysis is in

itself a miracle, for the theorem is concerned with inte­

gers, vihereas analysis deals with continuous and non-integral

numbers. For this reason much credit is c iven both Euler

and Dirichlet in the development of modern analytic theory

of numbers. In this branch of m^.thematics, analysis, the

mathematics of continuity is apolied to problems concerning

the discrete domain of the integer.

Many elementary proofs have been given for special

cases of Dirichlet's theorem, but it was not until l^k^

that Salberg gave an elementary proof of Dirichlet's theorem.

PART I

SPECIAL CASES OP

DIRICHLET'S THEOREM

CHAPTER I

ELEMENTARY RESULTS

Almost the first questions anyone might ask about

primes are, "How many primes are there? Are there an infi­

nite number?" Euclid was the first to give a proof that

there are an infinite number, and his proof is as signifi­

cant as it was when it was discovered over two thousand

years ago. Time has not mired its simplicity or beauty.

Let us look closely at it.

Theorem 1.1. There exist infinitely many primes.

Proof. Suppose on the contrary that there are only

finitely many, say p-., P2. ...t PT^. We form the number

(1) N = n p + 1.

i=l i

Then either N is a prime distinct from p- , ..., p, or is

divisible by a prime p. p cannot be any of p- , ..,, p- ,

for we should have from (1) k

p I N, P I TT p^ , i=l

and hence pll, which is impossible. Therefore p is a

pri.re distinct from j>-^, ..., p, , contradicting our hypothe­

sis.

Another way of stating this result is as follows:

There exists infinitely many primes of the arithmetic

progression 2n+l, which makes this result a special case

of Dirichlet's theorem.

Euclid's method of proof has become one of mathema­

ticians' finest tools. This method of proof will be used

in the next two theorems.

Theorem 1.2. There exist infinitely many primes of

the form ^n+3.

Proof. Suppose that there exist only finitely many

primes of the form ^n+3f say p, , p p, . We form the

number k

(2) N = 4 n P. - 1.

i=l

Then either N is prime and = 3 (mod 4) or N is composite.

Not all prime factors of N can be of the form 4n+l, for

otherwise N itself would be = 1 (mod 4). This is easily

seen by takineg the example x=4n-. + 1 and y=kn^ + 1 , t hen

xy = l6n-|n^ + 4n^ + 4n2 + 1 , or xy = 1 (mod 4). There

exists, therefore, at least one prime factor p = 3 (mod 4).

Again, p cannot be any one of p... p^. .... PT^, for if it

were, then from (2) k k

p I 4 Tf p. and p » 4 TT p. - 1 , i=l ^ i=l ^

which implies pi 1. This is impossible. Thus p is a prime

distinct from p, , ..., p, .

Now using the quadratic law of reciprocity of Gauss-

Legendre, we will prove the next theorem.

Theorem 1.3* There exist infinitely many primes of

the form 6n+l.

Proof. Suppose on the contrary that there are only

finitely many, say p , P2f ••.! PT * We form the number

(3) N = 12 ( n p.)^ + 1.

i=l

Then N = 1 (mod 6). Either N is a prime other than p]_,

P2» .... Pj or it is a composite. If N is a prime, we have

our contradiction, otherwise assume that a prime p divides

N. If p is two or three, then p I 12 and pi N, which implies

p divides one. This is impossible. Therefore p i 2, 3.

and pL N, so from (3). the congruence

3 x^ + 1 = 0 (mod p)

is solvable; i.e.

3 x^ = -1 (mod p),

3 (3 x2)= -3 (mod p),

(3 x)2 = -3 (mod p).

Thus

but by the quadratic law of reciprocity

Therefore

( ! ) • ^ ••

i.e., p = 1 (mod 3). o^ p=3k+l, where k is an integer, p

is odd since p ^ 2 and may be expressed as p=2m+l where m

is an integer greater than one.

So

P = 2m + 1 and p = 3k + 1

or 2m + 1 = 3k + 1,

2m = 3k.

But 2 l2m, so 2 1 3k. Since 2 t3, 2 I k, or k=2x where x is

an integer. So p is expressible in the form

p = 3(2x) + 1,

= 6x + 1.

Therefore there exists at least one prime divisor p of N

and p = 1 (mod 6). p cannot be one of the primes p^, ...»

p. ; for if it were, then from (3) k 2 ^ ^ 2

p 112 ( It p. ) + 1, p 112 (."n P ) ,

i=l ^ "- ^

and hence p I 1 which is impossible. Therefore there exist

infinitely many primes of the form 6n+l.

3y using this method, it is possible to prove a number

of special cases of Dirichlet's theorem.

CHAPTER II

ANALYTIC RESULTS

Euler gave a fundamentally different proof of Euclid's

theorem. This proof is done by analytical means, which is

the fundamental technique employed by Dirichlet.

In this chapter v:e will turn our attention to Euler's

proof and special cases of Dirichlet's theorem which may be

proven by analytic methods.

Let s be real and consider the series

(4) 2 -K n=l ^ ^

This series converges for s > 1, since

1 1

1 . 1 1 2 ^ •*• ""TS- 1 —oS: + ~Q? = oS-1

" ^ •*• "53" + ~ p " + "ys- 1 -^ **• - T ^ + "Tjjs- + - ^ - ifs-=ir-

etc, where we consider 1, 2, 4, 8, ... terms of the series.

It follows that the sum of any finite number of terms of

the given series is less than the geometric series

ls-1 - - 2 ^ **• " 4 ' ^ * " " 8 ^ + ... = i.(|)y-l

V7hich converges for s > 1. Thus by the comparison test the

series (4) converges for's > 1, and its sum is denoted the

Riemann zeta function ^ (s).

The zeta function, studied extensively by Riemann, is

8

fundamental in prime number theory. Its importance is due

to a remarkable theorem discovered by Euler which expresses

the function as a product of primes only [ ll] •

Theorem 2.1. For s > 1,

(5) ^ (s) = TT (1-p-^)"- . P

where the product is taken over all primes p.

Proof. Let

(6) P(x) = TT (l-p-^)-l, p<x

then since p > 2 and s > 1, we have

(7) • (1-p-^)-^ = 1 + p"^ + p-2^ + ... .

00

k=0 ^ .

Nox'j if each series (7) were absolutely convergent for all

P = P^. P2» P- * •••» ^-r— '^* ^^^ terms could be rearranged

in any order, and the rearranged series would converge to

the same sum [jTl, Each series (?) is absolutely convergent

since by the n root test C3l]»

-ks lim\lp-H)^'^^ = lim ^s k->cx) k->oo "

= P'^ < 1.

since p > 2 and s > 1. So

P(x) = TT d-p"^)"^. p<x

= n 2 p-^^ p<x k=0

= (1 + pi~^ + P-L"^^ + • • • ) • • • (1 + p / ^ + -2s ^ ^

Multiplying these series together, the general term resulting

is of the form,

Pl-^lSp2-^-2Sp^.a3s...p^-ars (^^^Q, n^>0, . . . a^>0).

Letting n = p ^ - p Po ^•••Pp . we have

(8) P(x) = 2 n" .

The number n will occur if and only if it has no prime factor

greater than x, and by the fundamental theorem of arithmetic,

it occurs once only. Hence in (8) the summation extends

over numbers formed from the primes p £ x.

These numbers n include all numbers < x, so that

n=l (x) x+1

and the l a s t sum tends t o 0 when x->flo • Hence

n=l x->»(x)

= l im n ( l - p " ^ ) " . X->£» p<x

= TT d - p " ^ ) " . P

Hence T (s ) = Vi (1-p"^) ( s > l ) . P

Theorem 2.1. may be regarded as the analytic analogue

of the fundamental theorem of arithmetic.

Theorem 2.1. is a special case of a more general theo­

rem which "Till be useful later. We recall that a function

f(n) is multiplicative if f(mn) = f(m)f(n) when (m, n) = 1,

and comoletely multiplicative if f(mn) = f(m)f(n) without

restrictions.

Theorem 2.2. If f(1) = 1 and f(n) is multiplicative,

then

10

(9) E ^M = TT (l+f(p)+f(p2)+...+f(p )+...) = n S f(p^)

n=l p p k=0

provided that either side is absolutely convergent. If

f(n) is completely multiplicative then (iO) TT Z f'(P ) =TT (l-f(p))-l.

p k=0 p

In both instances the product is taken over all primes p. CO

Proof. Suppose 2 f'(n) is absolutely convergent. n = l •

Let

(11) p(x) = n 2 f*(p^)-p<x k=0

00 00 Since 2 lf(n)l converges and 2 'f(p )• is a subseries of

n=l k=0

the series 2 ff (n)l , 2 If (D )l converges [ 4] . Since n=l k=0 00 ,

the series 2 f(p ) is absolutely convergent, we can rear-k=0

range terms and multiply the series together.

p(x) = n 2 f(p^) = '^ ( i+f (p)+f (p^)+-- - )

p<x k=0 p<x

Letting p^, Pp. .... P^ < x, we have P(x) = TT (1+f(p)+f(p2)+...)

P5.X

= Q + f (Pi)+f*(Pi)+- • -n iii+^(P2)+^(P2^^'*" • -H- • •

[;i+f(p^)+f ( p ^ ^ ) + . . 0

= 2f(Pi''^)f(P2^^)---f(Pr'''*)

where a's may take on any non-necrative integral value. The

summation is over all possible choices of a-^, a^. .... CL^.

Since f(n) is a multiplicative function and any two primes

p. and p. (1 5> j) are relatively prime.

11

P(x) = TT 2 f(P^) = 2 f(n) p<x k=0 (x)

where the last summation is taken over all n having no

prime divisor which exceeds x, and each such n occurs pre­

cisely once.

The summation 2 ^(n) includes all functions f(n) (x)

up to f(x), so that 00 CP

• 0<2 f(n) - 2 f(n) < E f(n) ; n=l (x) x+1

and the last sum tends to 0 when x->ao , Hence 00 00 00 , 2 f(n) = lim 2^M = lim TT E f(P ) = TT 2 f ( p ^ ) .

n=l x->oo (x) x->oo p<x k=0 p k=0

Now l e t

(12) TT 2 ^(P ) ® absolutely convergent, and consider p k=0

(13) Q(x) =17 2 f(P^).

p k=0

The absolute convergence of (13) follows as a corollary of

the convergence of (12) C5D« Since the series in (13) are

finite and Q(x) is an absolutely convergent product, we may

multiply the series together and rearrange factors of the

product without affecting the value of the product [yT]. Q(x) =T\ 2 f (P ) .

p k=0

= TT Cn-f(p)+f(p^)+---f(p^)D • P

Multiplying the series together, the general term resulting

is of the form

f(Pl l)f(P2''2)f'(P3'' )---. (0<ai<x; i=l, 2. 3. • • • ) .

Noting that we are dealing with multiplicative functions and

12

(p^tPJ = 1 where i j- j, we have

A number n will occur if and only if it has no prime factor

with an exponent greater than x, and each n occurs pre­

cisely once. Hence Px ^

Q(x) = 2 f'(n). F^ = (p-j_P2P3...) .

Thus ^ 00

0 < 2 f(n) - Q(x) < 2 f(n). n=l F +i

The last sum tends to zero as x ->00 . Hence 00 GO V 00 V E f (n) = lim Q(x) = lim IT 2 f (p ) ^ U E f'(P )• n=l x->oo x->oo p k=0 p k=0

Now v:e wish to consider the second part of the theo­

rem (10). Since we are dealing with completely multiplicative

functions, 00 00 ]^

-n 2 f(P ) =TT 2 Criv)3\ p k=o p k=0

= TT (1 + f(p) + [If(p)I] + &(P)I] +•••). P

Now we shall show that 2 3 -1

1 + f(p) = &(P)I] + Cf (P)I] + ••• = Cl-f(p)Il when

lf(p)l < 1. 2 n-1

Let S„ = 1 + f (p) + &(p)Il + ••• + Cf (P)I1 2 n-1 n

then f (p)S^ = f(p) +Cf (P)II + ••• + &(P)I] + &(P)I] • n

Subtracting, D-^(p)IlSn = 1 - &(p)Il *

i-[:f(p)J °^ ^n = l-f(p)

n Now we need to show that lim df (p)]] = 0. To prove this we

n->00

13

must show that given € > 0 we can find N such that n

I Q(p)D - 0|< € for n > N.

This result is true for f (p) = 0; hence we consider f (p) 5 0.

Now |Q(P)I] 1= |f(p) < €

when n Inlf (p)| < In £ , or n > (In £ ) / (In I f (p)l ) = N

Qsince if f (p) < 1, In f(p) is negative 3* We have there-n

fore found the required N, and lim Qf(p)] = 0. Thus 2 n->QO

1 + f(p) +Q(p)I] + ••• = lim S^, n->co

^'^ i-f(p) , r^p),

Thus we have completed the proof of Theorem 2.2,

In order to prove that there are an infinite number of

primes, Euler proved that the series 2 1/P diverged. P

Theorem 2.3.

s->l" 0 p pS

wiTEiB the sum is over all primes p. There exist therefore

infinitely many primes.

Proof. From (5). we have

iis) =Tr ( i -p -^ )"^ , s > 1. ^ p

Now taking the logarithms of both sides, we have

log llis) = log TT (1-p-^)"-^. ^ P

= - 2 log (1-p-^). P

But since s > 1 and p > 2, we have -l<p~^ < 1; therefore we

can expand log (l-p*^) as log (1-p-S) = - ^ 1 ^

k= 1 k p ^ .

14

Hence log ^ (s) = 2 S - j ^ ^ p k=l ^P

00

p k=2 ^P

= 2 VP^ +2 ^ 1 ^ . p p k=2 ^

Now we need to look at the behavior of log T (s) as

s->l ^. In order to do this, we will first look at the behavior of ^ (s) as s->l"*"0.

We can write )^ (s) in the form

^ (s) = 2 n - ^ = J"x-^dx + § j''"' (n- -x-S)dx. 1 1 I n

Here \ x~ dx =sn"

since s > 1. Also

n-s - x-s = j' 0 < n-^ - x-^ = \^ st"^'^ dt < nl n

if n < X < n+1, and so

0 < 1"*^ (n-s - x-^) dx < la : ^n

and m r n+1 _ _« 2 I (n'^ - X ) dx is positive i V n

and numerically less than s2 n~ .

Hence

^ (s) = + 0 (1) =(^) (l + 0(s-l)j ,

and log ^ is) = log s^x + log (l + O(s-l))

= los s ^ + O(s-l).

Therefore Log t (s) ->«> as s -> 1, so if we could prove that 2 2 TT^S remains bounded as s -> 1, then we would

p k=2 ^P have proved the theorem.

1 5

I t R(s) =2 £ r p ^=2^^"^

then JR(s)| < 2 2^ Ij^l p k=2

1 2 2 ^pte p k=2 -

1 *2 2 p p k=2 ^

ri

^^t 2 5te = (i-p-2)"i. (s > 0)

k=0 ^

CO ^ 1 1 - , 1 P-^^

-2s or IR(S)I<|2 f- s

P ^

5. 2 - T-s 2 P •

: 1 ^ -2s 2 1-2-2

< r— . 2 n n=l

o. -2s Now looking at the behavior of the series 2^=1 n" as

s -> 1 , we see

lim 2 n" = 2 "H " • -he series on the right-s->l" ^ n=l n=l

hand side of the equation is a hyperharmonic series, and

converges \J?'2 • Hence as s -> 1 , R(s) remains bounded

and

im , . 2 ^P^ -s->l+0

> cx>

Corollary 2.3. As a simple corollary, we infer that

2 /p diverges. P

Proof. Suppose 2 1/P converges to a. P

Then for s > 1,

16

2 l/P < a. P

and this contradicts the statement of the theorem. Thus

2 1/p diverges, P

We now wish to modify Euler's argument to prove a spe­

cial case of Dirichlet's theorem. To do this, we introduce

a character X(n) and a series

L(s.X) = f % i n=l ^

We make the following requirements of X(n):

(1) X(n) must be completely multiplicative; i.e.,

X{m)X{n) = X(mn),

(2) X{n) depends on the residue class modulo k to which n

belongs, and

(3) X(n) = 0 if (n,k)> 1,

Using these modifications, we will prove the following

theorem:

Theorem 2.4.

->a?. Ii F - i->i ° pslTmod 4)P^ S'

There exist, therefore, infinitely many primes in the pro­

gression 4n+l,

Proof. Let us define our character X'(n) as follows: . ,,(n-l)/2 ^ (-1)' for odd n,

X(n) =

0 for even n.

The second and third conditions for a character are obviously

displayed by our function. We need only to show that our

function is completely multiplicative. If either m or n or

17

both ra and n are even, then X(m)) (n) = /(mn). Now consider

both m and n odd,

X(m) = (-1)^^-^)/^, ^

X(n)= (-l)^^-^^/^

X(m) X(n) = (.l)(-l)/2(.i)^-^)/2^

and X(mn) = (.i)(^-l)/2.

X(mn) = X(m)X(n) if the exponents have the same parity.

To show that this is the case, we need only to show that

the difference of the two numbers is an even number,

mn-1 (m-l) + (n-l) _ mn-m-n-H _. (m-1) (n-1) 2 " 2 •" 2 2

Since m and n are both odd, both (m-1) and (n-l) are even.

Thus (m-lVn-1) is even. Hence X(n) is a completely mul-2

tiplicative function.

Consider now for real s the series

L(s, X) = § iLLnl n=l n^

1 i + 1 1 + = 1 - 3 s + s " s + • • • •

n=0(2n+l)s ^

From the elementary theory of infinite series, vie now

show that this series converges for s > 0, and converges

absolutely for s > 1, Let

^n = (k!n-H)s

The alternating series 2 (-l) a (n=0, 1, 2,... and a^ > 0)

converges if (1) an>an+i (n=0, 1,...), and (2) lim a ^ = OQ 8]],

If s > 0, this is the case since 1 ^ 1

(2n+l) 2 - L2(n+l)+in . and lim 1 = 0.

n->00 (2n+l)^

18

The series 2 Tvn-H)^ converges absolutely if the n=0

CD series 2 (2 +1)^ converges. This is the case for s > 1

since the series 2 —=^ (s>l) converares and each term of the T nS n=l ^

00 series 2 /^ -, \o is less than or equal to the corresponding

^2. T2n+1) s CD

term of the series 2 ~s • Thus by the comparison test, n=l ^

the series 2 , vo converges absolutely for s>l FQ"] . (2n+l)^ ~

Since X(n) is completely multiplicative, and L(s,X) is

absolutely convergent for s>l, we have from Theorem 2.2,,

I-L(s,x) i n Q - ^%I T"^ . odd p P'

Taking logarithms, we get

Log L(s.X) = log n C l - ^ n '^. odd p

= - 2 log.ci - ^ : • odd p P

Since s > 1, p > 2, and 'X(p)l < 1, vie can expand the loga­

rithm as

log Ci - Mia = - S | M . pS =2. 'P

Thus

log L(s,X ) = 2 I fM^. odd p k=l '-P*^

Odd p P k=2 ^P

_ K X + ^ ^ X(p^)

Odd pP odd p k=2 P

19

1 - "^ 1 + R, ( s ) . p=l(mod 4) P ps3(mod 4) P

where we c a l l e d the t h i r d sum RAs).

As be fo re we have

Hi(s)l< X 2 4 i # odd p k=2 '^P^^ ,

k' < i 2 2 - ^ ^ odd p k=2 P

ca CO

^3-

but

so

odd p k=2 P

^ - k r = d - p ' ^ ) " - ^ fo r s > 0. k=0 P^^

5* 1 _ ^

or

Now s

Bi(s)| < * 2 odd p

< i 1 - 2 l-2"2

< -1 1

ince the series

p-2s 1-p-s

£ p-2^

n=l

n=l converges as s -> 1 ,

Rl(s) remains bounded. To isolate the primes p=l(mod 4),

vje add log ^ (s) where

log ^ (s) = 2 ^ ^ + R(s). P P

= -Is- + 2 -^ + s(s) . ^ odd p P

= 4^ + 2 4 ^ + 2 - ^ + R ( s ) . ^ p=l(mod 4) P p=3(mod 4) P

20

log L(s.x) + log t (s) =-is- •*• 2 2 ; ^ + Rl(s) ^ ^ p=lTmod 4) P

+R(s).

We know that log t (s) ->oo as s -> 1 and the terms

R-j_(s), R(s), and 2 all remain bounded as s -> 1 . Also,

L(s,X) -> L(1,X) a convergent series. Our theorem would

be proved if we know that L(l,x) ? 0. This can be seen as

follows

LCI, X) = (1 - ) + ( - ) + ••• > 0.

thus proving the theorem.

PART II

THEOREMS NECESSARY TO THE

PROOF OF DIRICHLET'S THEOREM

CHAPTER III

CHARACrERS

In order to prove the general case of Dirichlet's

theorem, we need to develope the general theory of charac­

ters. In this chapter we will develope the theory of

characters for an Abelian group G of order h and then apply

it to our particular needs.

Recall we made the following requirements of a char­

acter:

(1) X(n) must be completely multiplicative, (2) X(n) de­

pends on the residue class modulo k to v:hich n belongs —

the most important being the reduced residue class, and

(3) X(n) = 0 if (n, k) >1,

If X(a) = 0 for every element in our group, then

X(a) satisfies the multiplicative property, but we agree

to exclude this trivial case. Also the multiplicative

property is satisfied if X(a) = 1 for all elements in the

group. V/e shall call this the principal character and de­

note it X]_(a). It is not obvious that there exist other

characters, but we shall show for an Abelian group G of

order h, there are h distinct characters.

A few properties can be proved directly from defi­

nition,

22

23

Theorem 3.1. If e is the identity of the Abelian

group G, then X(e) = 1.

Proof. We have

X(e) = X(e2) = (e):! .

Hence X(e) = 0 or X(e) = 1. If X(e) = 0, then X(a) =

X(ae) = X(a)X(e) = 0, for each a in the group, and we have

excluded this case.

Theorem 3.2. X(a) is different from 0 for all a in G.

Proof. Suppose on the contrary that for some a in G,

X(a) = 0. Then X(e) = X(aa-l) = X(a)X(a"^) = 0. Thus if

X(a) = 0. then X(e) = 0, which is a contradiction to Theo­

rem 3.1.. so X(a) is different from zero for all a in G.

Theorem 3-3. X(a) is an h^^ root of unity; i.e. X(a)

h satisfies the equation X (a) = 1 for all a in G,

Proof. Let a ^ G. Since G is a group of order h,

a = e. Thus

X^(a) = X(a^) = X(e) = 1.

We now state a fundamental theorem of Abelian groups

which is necessary in our next theorem. A proof to this

theorem can be found in Theory of Groups by Marshall Hall CloU'

Theorem. Let G be an Abelian group of order h; then

there exist in G elements a , a^ a^ of order r-,, r^,

..., r respectively so that every element g of G may be

written uniquely in this form

g = a-, la<, 2 ... a^^^ where G < k . <r, - l ( i = ^ 1 2 m — i — i

l,2,...m) , Moreover ^1*^2'**^m ~ *

24

Theorem 3-4. There are precisely h different charac­

ters for a group G of order h.

Proof Let an, ,.,, a be a basis from G and let r, J- m j

be the order of a. (;) = 1,2,... ,m), We define

(1) X(aj) = Pj (j = l,...,m),

th where p. is the r. root of unity and may be expressed as ^ 2i£l

(2) p. = exp ^1 * i 0 < n.<rj - 1.

ki kr Also If g = a- .,, a , let

^1 •. ^1 r (3) X(g) = p{^^ ... p_-^ .

Then X so defined Is a character, for if a and b are two

elements of the group G, we have

SI S2 Sm a — a a ) .,, a J. 2 m

and

ti t2 tn b = a-L -Lag ''...ajj ^ .

Hence from (3)

y/ N ^ SI S2 ^ Sm X(a) = pL P2 •• -Pm

X(b) = p-L • P2 •••Pm^ m

and X(ab) = X (a ^ "' ,, .a/^""^^).

Suppose that s^ + t^ = u. (mod r^), 0 < u^ < r^ - 1 si+ti kiri+ui , ,

then a^ = ^i ^^^ ^^^°^ 2)

= p .

= p"i.

Thus X(ab) = Pi Pa"^ • • • Pm' '" !

25

and X(a)X(b) = p^^^^^Sa'^**^. • •p/"'**'".

I n

But pj = pj i f 1 = n(mod r ^ ) , and hence X(ab) = X(a)X(b).

The numbeir of poss ib le choices for n . I s r . and hence the

t o t a l number of poss ib le choices for X(g) = X ( a . ^ ^ . , , a ^'^) ± m

is ^1*^2" *^m " * ^"^^^^OYQT, no two distinct choices lead

to the same character. For if X, and X are two characters u V

defined by

X (a ) = exp 2ni . ^ 0 < u < r, - 1, " J r j J " " j " " j

and

X (a ) = exp 2TTi . 0 < v, < r, - 1,

then for some j, u ^ . Thus there exist an element g

in G for which

X^(g) 5 X^(g) ,

the element being namely g=a.. There are, therefore, at

least h characters. On the other hand if X is a character,

then X(g) = X(a^ • )*''X(a ° ). Each of the factors is an

th r. root of unity, and so X(g) must coincide with one of

the above choices. There are, therefore, exactly h distinct

characters.

Characters may be combined in the following way: Let

X and X be two characters on the group G, We define for g

in G

X X (g) = x(g) X(g),

Theorem 3»5- Under this rule of composition, the

26

characters of a group G form themselves an Abelian group.

Proof. First, because of its usefulness in the rest

of the proof, we will show x X (g) = X x (g). Let

x(a.) = p where p. = exp(2lll . n,) J J J r . J

where (j = 1, ..., m) and 0 < n. < r - 1 ;

and let X (a ) = q. where q. = exp (2ni/r. . m )

were (j = 1, ..., m) and 0 < m. < r - 1.

Also if g = a, - •••a let 1 r

/ \ ^1 ^r x(g) = p^ ... p^ and x/f \ ^1 ^r

x(g) = qi .-. q^ . Then

X X(g) =x(g) X(g),

. ki ^r N / il kp. ' = (p^ - ...P- ) iq-^ - ...qp ^ ) .

= exp 2TTi C ( ^ + ... + !il±r)^(5i^ +...+!^):] • 1 Tp ri r^ -•

= exp 2TTi L ( - ^ + ... + _£_^) +

(21^ + . . . + ^v^) :i

= (q -'l .-. q/^)(Pi^^--.Pp^^^) '

= X(g) x(g) = X x(g)

The set of characters is closed. Let U/= x X, Then

l^(ab) = X X(ab) = x(ab) X(ab)

= x(a)x(b)X(a)X(b).

= x(a)X(a)x(b)X(b),

= xX(a)xX(b),

= li/(a) "tp'(b). Thus lif is a character. The

associative law nolds for characters. Let a, 3, and y be

27

three characters defined on G such thata-3(g) = x(g) and

PvCg) = X(g). We Wish to show that aX(g) = xyCg).

aX(g) = a(g)X(g) = a(g)aY(g) = a(g) 3(g) yCg).

XY(g) = x(g)Y(g) = a3(g)Y(g) = a(g)3(g)Y(g).

Thus the associative law holds for characters. The princi­

pal character X^ acts as the identity since

XjXis) = XX3_(g) = X(g),

If g is an element of G and

X(g) = Vi • .. .Pj. .

let {jjie) = Pi'^'^.-.p/^^ .

Then X(g)i^(g) =i^(g)X(g) = 1 = X^(g).

Thus for a character X(g), there exists an inverse li/(g).

Hence we have shown that the characters of a group G form

an Abelian group.

Theorem 3.6.

h if X = X. ,

> 2 ^(g) = .

g6 G |0 if X 5 X^,

the sum being for a fixed character summed over all elements

of the group.

Proof, If X = X3_, then X^(g) = 1 for all g in G

and so 2 ^1(s) = ^ since the order of the group G is h. g€ G 1

If X 5« X-j_, then let

S = 2 ^(g)-

Suppose that a is an element of G for which X(a) 5 1. Such

an element must exist, for otherwise X viould be the principal

28

character. As g ranges over the elements of G, then so

does a«g once and only once. Then

S = 2 > (g) = 2 5((ag) = 2 X(a)X(g) g€ G g£ G g^ G

= X(a) 2 X(g) = X(a) S, g€G

Since S » X(a) S and X(a) 4 1, then S must be zero. Thus

we have completed our proof, i^

VJe now prove the so-called dual of this theorem.

Theorem 3 . 7 . fh i f g = e ,

2x(g) =. X |0 o t h e r w i s e . \

the sum being taken over all characters X for a fixed

element g of G.

Proof. If g = e then by Theorem 3.1. X(g) = 1 for all

X, and since there are precisely h different characters for

the group G, then

2 5i(e) = h. X

Suppose g 5 e; then we show that there exists a character

X such that X(g) 5 1. To this end we let

g — a-| ...a ^1 ^ im

m since g 5 e; then for some j, k. 0 (mod r^). We define

X(a.) = exp and

X(a^) = X(a2) = ••• X(a 3_) = X(a^^^) = ••. = X(a^) = 1,

Then, as we saw above X defines a character and X(g) =

"9^^^ ? 1 Esince k. 4 O(mod r .) and p. .1 = exp iULi 4 \ \ . J «J J J I* <

Now we let

29

S =2 ^is) •? and since by Theorem 3.5, x X ranges over all characters

whenever X does, we select x(g) 4 1, then^

S = 2 3f(g) =2x(g) X(g) = x(g) 2 ^ ( g ) X X X

= x(g) S

and since x(g) 4 1 and S = x(g) S, then S = 0, Hence we

have completed our proof.

yrl t'r . •r^:r .

CHAPTER IV

THE GENERAL THEORY OP DIRICHLET

SERIES

• = * , -

A Dirichlet series is a series of the form

(1) ? a(n) n=l

Though Dirichlet used only functions of real variables in

his proofs, we shall generalize this by letting s = a + it

and by letting a(n) be a sequence of real or complex num­

bers. Here we have followed the standard notation of

denoting the real part of s by the Greek letter a and the

imaginary part by the Latin t.

The theory of Dirichlei series involves many delicate

questions of convergences; here we will establish a few of

these results which will be useful in further development

of our sub.iect.

Theorem 4.1. If the series (1) converges for

S- = a +it , then it converges for all s for which o>a_, o o o o

Moreover, the convergence is uniform inside any angle for

which larg (s-s^)! < n/2-6 for O<6<TT/2.

Proof, The proof is based on a lemma for partial

integration of Stieltjes integrals which we state in the

follovring form.

30

imMiN.

31

Lemma 4.1. Let x > 1 and let 1 (x) be a function vxith

continuous derivative for x > 1. Let Six) = 2 G(n) n<x

where C(n) are r e a l or complex numbers. Then

(2) 2 C(n) ^^(n) = Six) XJJix) -• j ^ S ( t ) ' l / ; ' ( t ) d t .

Proof of Lemma 4.1. Let k be an integer such that

k < X < k+1; then in this interval vre have

tx] k (3) S(x) = 2 C(n) = 2 C(n) = 2 C(n) = S(k).

n<x n=l n=l

where QT] denotes the g r e a t e s t in tege r in x .

We define S(0) = 0., and we get

2 C(n) \ljin) = 2 C(n) ihin) n<x ^ n=l ^

= 2 C(n) -[bin) n=l ^

But s ince

S(n) = 2 C(x) x<n

and S (n - l ) = 2 G(x), x<n-l

we have

S(n) - S (n - l ) = 2 C(x) - 2 C(x).

x<n x<n-l

= C(n).

So r ep lac ing C(n) by S(n) - S(n- l ) and proceeding x\*e have

2 C(n) l//(n) = 2 Cs(n) - S(n- l ) 2 ll/(n) Kx ^ n=l ^

32

2 [IS(n) ^ ( n ) - S ( n - l ) xfjin) J . n=l

k k = 2 Sin) xjjin) - 2 S ( n - l ) i/;(n) ,

n=l ^ n=l ^

k k - 1 = 2 S(n) \ i ;(n) - 2 S(n) i / /(n+l) ,

n=l ^ n=0 ^

on changing the index i n the second sum. Now proceed ing

we have

k k -1 2 C(n) z//(n) = 2 S(n) t/;(n) - S(0) xj/il) - 2 S (n ) i / / (n+ l ) n<x n=l n=l

= 2 " S(n) W;(n) + S(k) W;(k) - 2 S (n ) \ l / (n+ l ) n=l ^ ^ n=l ^

= k - 1 2 S(n) [ : \^ (n) - \pin+l)2 + S(k) i / ; ( k ) .

n=l

By adding and s u b t r a c t i n g S(k) xjjix), we have

k-1 2 Oin) xhin) = 2 C l/ (n) -l/;(n+l) S(n)n + n<x n=l

s(k) H \/;(k) - i/;(x):] + s(k) \ j j i x ) . , I C n+1 , "I

We now replace \p in) - XjJin+l) ^y - - n r ^^^^^ ^^

Jx "? ^1/; (t)dt and obtain

k-1 r n+1 » 2 G(n) W;(n) = - 2 S(n) ] W/ (t)dt n<x ^ n=l ^ n r "" (X •

- S(k) ) ijj (t)dt + S(k)'l//(x).

Since S(x) = S(k) from (3) and S(n) is constant on the

intervals n to n+1 and k to x, we may treat S(n) as a

33

constant and let S(x) = S(k) to obtain

2 C(n) \hin) n<x ^

^-1 /n+1 , t (y. I - 2 ) S(t)i/; (t)dt - )l Sit)xlf (t)dt n=l n ^ k r

+ S(x)V^(x),

= s(x) \/;(x) - 2 i"""- s(t) \//'(t)dt n=l ^

- J s(t)i/;*(t)dt,

= S(x) l/;(x) - C J S(t);/;*(t)dt

+ $2 S(t)i^*(t)dt +••• +

ik-1 s(t)i/;'(t)dt:] - i^ s(t)i/;'(t)dt

= S(x)i|;(x) - ^^ S(t)l/;'(t)dt,

Thus we have completed the proof of the lemma and now re­

turn to the proof of Theorem 4.1,

We put 00 GO ^ a(n) = ^ a(n) ^-^ --«=! ^ - n o

a(n)/_s,^ ././„\ _ ^-(S-SQ)

^^3_ ns JTn n'=*o ns-So n=l

= X and C(n) = " ' " V n ^ o , •^^(x) =

then

S(x) = £ C(n) = E ||^ n<x n<x

and xj/'ix) = -(S-SQ) X-^-'^'-^O' .

Now

34

M , . M N Z aiSi = £ aM _ ^ a(ni

= 2 a M . -1 . 2 a(ni . 1

Now using lemma 4.1. we obtain

M z ^ n=N+l

I|i = S(M) ;//(M) - ij S(t) ;/;'(t)dt

- S(N) l/;(N) + J' S(t) l/;'(t)dt.

But since ^(M) = M-(S-SO) ^ r^.j -(S-SQ)

and, l//*(t) = -(S-SQ) t"^-(s-So) ^ ^ ^^^

2 § ^ = S(M)M-^^-^O) + (s.s ) (M s(t) t-l-(s.So)^t n=N+l n " 1

-S(N) N-(s-So) . (3_3^) jN g(^)^-l-(s-So)^^^

Adding and subtracting S(N) M"^^"^o) and combining the

integrals, we have

M 2 ^igi = CS(M) - S(N);] M"^^'^o^

n=N+l n / ^ ^ ^ •+ S(N) Qr^^'^o^ - N- ' o-'I]

+ (S-SQ) i,J S(t) t-l-(^-^o)dt. o N

But S(N) QM'^^'^O) « N'^^'^o^]

= -(S-SQ) ijj 3(N) t"^"^^-^o^dt,

so 2 ^ ^ = CS(M) - S(N)I| iC^^'^o) n=N+l ^

- (s-So) \l S(N) t-l-(^-^o)^t.

fw.r.,v--' 35

+ (s-s^) iJJ S(t) t-^-(^-^o)^t, " N

= CS(M) - S(N)3 M"^^'^O^

+ (S-SQ) iVs(t) - S(N):] t-^-^^-^o^dt. N

00 Since we have assumed the 2 a(n)/n ° converges, then for

n l € > 0 there exists N. such that for all x > N > N-, we

IS(x) - S(N)| < £ ; thus

+ (s-s^) i: Cs(t) - S(N)J t-^-^^"^o^dtl. N

<|S(M) - S(N)| • I M-^s-So) I

/Ml + S-SQ )j^|s(t) - S(N)

<iS(M) - S(N)| I n'^^'^o)

.-l-(s-So) dt ,

by using S(x) - S(N) N

< £ . •

dt .

For a > JQ, vie have

j«|,-l-(s-So)|,, . S« t-^-(°-o)dt = N-(' -°o)-H-'°-°o) N'" "• -N

Now substituting this result in, we have

M 2

n=N+l

a(n) ns

<ls(m) - S(N)

+ 6 Is-s,

. -(s-so)

^vj-(g-go),M"'^^"^Q^

Q-Co

36

We know (S(M) - S ( N ) | < € s i n c e M > N, a n d t h a t | M " ^ ^ " ^ O ^ I =

M-<°-°o) , 0 .

2 ^ < € M-^^-^o) +|s-So)r-N-(cJ-ao).j^-(a-ao)-,\ ln=N+l n " ^ cf-oo '- . - ' /

Now s i n c e j s - s j >/Re ( S - S Q ) | = a-a^ ,

M-^^-^o) <lfl!o! M-^''-°°^ . and - a-Oo

^ inS+1 "^1 ' ( S ' ' ' '^CN-^-o) -M-^^-^'

|s-sc| N--( - o) < £ 1° q N

M O-OQ

s Since 6 is arbitrary, (3) implies that 2 a(n)/n may be n+1

made arbitrarily small for a fixed C!>OQ. Hence the series

converges. Now we need only to show that the series con­

verges uniformly inside any angle for which

larg (S-SQ)I <• n/2 - § for 0 <§< n/2.

We know that

arg |s-So|= arc cos jg^s^ ' .

or

COS arg |fe-So) = -j- - j .

so

/s-s I- ^-^o '^ ol - COS arg IS-S-QI .

Hence

37

I S-Sr>l |cf-^ol "

a^a a. COS arg )s-s

1

I • T ^

= cos arg |s-s-| ,

But

cos arg 1S-SQ) — cos |arg s-s )

since arg J(S-SQ)|<^

and js-s jis real.

so

Is-s-s a i^^-^oi — cos Iarg S-SQI

1 = cos(n/2-6) .

1 sin6

Hence from (3) we have

M a(n) 2—^ n+1 ^ ^

(z

slnS

and the convergence is thus uniform inside the angle.

Corollary, If

00

2 n=l

a(n) n

converges, and

f(s) = 2 §l§i ntl ^ ^

then

lim f(s) = 2 ^ ^ S->So n ^

where s->So through values in the region of uniform con-

vergence described above.

38

Proof. Since we are dealing with a uniform convergent

series, under our given hypothesis

lim f(s) = lim 2 ^ ^ s->s, s->s, n-

= 2 lim i ^

= 2^^ A Dirichlet series may converge for all values of s,

or for no values of s, or may converge for some values and

diverge for others. For example, the series

00

1 n In^ converges

for every s, for by the ratio test, vre have for every s

s lim n->a) (n+1):(n+1)" nln^ = lim n

n->oo (n+1)(n+1) = 0 ;

on the other hand, the series

? n'

is always divergent, for by the ratio test

lim n-x»

(n+1 Tn+T

.' . n.' = lim n->aD

n TE+n =± = 00

for every s.

Theorem 4.2 The series (1) either converges for all

values of s or diverges for all values of s, or there exists

a line in the complex plane, a = a, such that (1) con­

verges at all points to the right of this line and diverges

at all points to the left, VJe call a the abscissa of con­

vergence •

T>

39

Proof, Prom Theorem 4.1, we see that if the series

(1) converges for SQ^a^+itQ, it converges for all s for

which a > OQ, We now divide the real numbers into two

classes R and L. Y belongs to R if series. (1) converges

for all values of o > Y. Otherwise, Y is placed in L, By

Theorem 4,1, every member of R is to the right of every mem­

ber of L, and there are members in both classes. Therefore

a Dedekind cut is defined, and the real number defined by

that cut is a, the abscissa of convergence. Thus the re­

gion of convergence is a half-plane to the right of a = a.

On the line a = a series (1) may converge or diverge.

Theorem 4.3. The series (1) whose abscissa of conver­

gence is a, represents an analytic function in the half

plane a > a.

Proof. Let s_ = a^ + it^ and suppose that a_ > a. o o o o

Then there exists a neighborhood of s^ lying in a region

of uniform convergence. Since each term of the series is

analytic, it follovjs that the sum of the series is an ana­

lytic function, for the series converges uniformly in the

prescribed neighborhood of s^ CllH* Theorem 4.4. If a(n) > 0, and f(s) is the function

00 determined by 2 a(n) n , the f(s) has a singularity at

n=l s=a, the abscissa of convergence.

Proof, . The proof is sinilar to its power series

analogue. Let a-, > a and expand f(s) in a Taylor series

about a . We get

40

f(s) = 2 tillio^L. (s-ai)^ k=0 ^! ^

If a is not a singular, then the interior of the circle with

centre a^ and radius a^ - a lies in the half plane a > a,

and has only regular points on its boundary. Hence the

radius of convergence r is greater than a^ - a since the

circle of convergents of the power series must have at

least one singularity on its boundary. The idea of this

proof is that the circle of convergence goes to the left

of a, and so there must be a point to the left of a = a,

at which the Dirichlet series converges. This contradicts

the choice of a. Here are the details: choose € > 0 so

that the radius of convergence of the power series

r > a - L - a + 6 > a - j _ - a . Then the Taylor expansion of

f(a - € ) is;

k=0

This series converges. Since o-^ is in the circle of con­

vergence of the power series, the series f(a-j ) =

00 2 ^^^' ^ay be differentiated term by term, and the

n=l n 1 derived series converges and represents the derivative of

the sum of the original power series Cl2]] . Thus

f(ai) = 2 ^ n=l

CO

(4) f(a-6) = I f^^Uo^) ^i=^iS^

f '(ai) = 2 afai (-logon) n=l 1 n ^

4-1

f"(o^) = f a(n) [:-log (n)i2 n=l nc l

f(^)(aO = f a(n) C-logenH^ n ^ nc l

Now substituting this result into (4) we have

ria^) = g Ia=£-lii I (-10 n)^ a(n) k=0 ^' n=l " -

= 5" 5" C(a-£-ai) (-log n)2 a(n) k=0 n=l kJ n^l • .

= 2 2 C(gl-a- €)(log n)n^ a(n) k=0 n=l 2 n l •

Since by hypothesis a(n) > 0 and o-^ - CL +€ > 0, all terms

involved in the double sum are non-negative; we therefore

have an absolutely convergent series and may interchange

the order of summation Cl3ll.

f(a-£) = 2 2 C(gl- - )(log n)J a(n) n=l k=0 k! n l

_ ^ a(n) f Cdos n)(ai-a+£)n^ •" 4^ y,al ^ TTj

n=l n k=0 ^. V/e now have, s ince CO r- . , ^j^\"[^

S* L d o g n)(a2^-a+€)J k=0 k]

ex £ ( l o g n)(ai-a-i^)>n' '^"^-"'^ ,

0 0 , X ^

f (a-e) = j ; ^ 1 : ; ^ . n-cJi+a-^ ,

„^j,<'

42

CO = 2 jg] nei H ^ .

Thus the series with s=a-(£ converges contradicting the

fact that a is the abscissa of convergences. n

Theorem 4.5. Let S(n) = 2 a(k). If 3> 0 and S(n) = k=l

0 (n^), then the abscissa of convergence a is < 3. Thus

if Y is the lower bound of numbers 3 for which S(n) = 0(nP),

then a < Y«

Proof. Since

n n-1 S(n) - S(n-l) = 2 a(k) - 2 a(k) = a(n).

k=l k=l

we have for M > N

M M 5 a(n) = 5; S(n) - S(n-l) n=l\[ nS n=N n^

M M = y ^M ^ S(n-l)

i N ""n^ ^N n^

M . . M S S(n) 5* S(n-l) S(N-l) rfeN""? n=N nS - N^

M _, . M ^ S(n) S S(n) S(N-;1) ^ n^N "^I^ ^ N (nvl)^ ' " " N ^ ,

< S(n) 4 S(^). . S(M) S(N-l) - ^ N " ^ ^ '"n=N Ti TiTys •" Ti+rys - NS ^

4 ^rn^ r 1 1 \ . S(M) S(N-l)

( n+1 ^ ^ •} 1

^ ° 2 • > n 3E^ = -i^ - T ^ s . so

43

M 2 ^ n=N n""

M 2 n=N

S(n) n+1

n dx S(N-l) ^ S(M)

(M+1)S

and since S(n) is constant over the interval n to n+1.

M , . M , n+1 2. 4 ^ = s 2 J n^N n n=N n

S(x)dx - S(N-l) , S(M) xs+x NS "*• (M+1)

= s N+1 ( ^"^ S(x)dx +. . (M+1 S(x)dx N M sTT

S(N-l) S(M) NS IM+TT^

= sf M+1

• N

S(x)dx S(N-l) . S(M) (M+l)S

= 0 , ( " 1 IS(x)l dx ] ^ O(N^) ^ O(M^)

0( isl i H^^ ^ ^ ^ ) ^ 0(N^-^) + 0.(M - )

I M+1 = 0 I (si N x-^-l+^ dxy+ 0(N^-^) + 0(M^-°) ,

= 0 ( |sl^C(M-*-l)^-^-N^-^: )+ 0(N^-^)+0(M^-^).

If a>P, the right-hand side -> 0 as N ->oo , thus proving

that the series converges.

Since the Riemann zeta function is an important exam­

ple of a Dirichlet series, v:e will now develop one of its

important properties.

Theorem 4.6. The function

oo (4) ^ (s) = 2

n=l n' where s = a+it is analytic in the

half plane a > 0 except for a simple pole at s = 1 with

residue 1.

44

Proof.

S(n) = 2 a(k) = 2 1 = n, k=l k=l

>

By Theorem 4.5. If 3 > 0 and S(n) = O(n^), then

a is < 3. If vje choose 3 = 1, then these conditions are

siatisfied and a < 1, but (4) converges for a > 1 and di­

verges for a = 1 so the abscissa of convergence a must

equal one. V/e now wish to effect a continuation of the

zeta function by the use of the Euler-Mac Laurin formula

which follows.

Lemma 4,2, Let \h{x) be defined and continuous for

n < X < m, where m, n are positive integers; then

^0,) = i : i | i . % ) , $^^u),

+ ')l ^'(x)(x-Cxn-l) dx.

Here CTQ denotes the greatest Integer in x.

Proof of Lemma 4.2. Starting viith the last term on

the right, we have

i^ XJj'ix) x-CxH-t <ix = in^*(x) . X dx - i^"^'(x) K dx

- h )^IJJ (x)dx.

(m I ( n + 1 I (n+2 , i Now ) ' \jj ( x ) x d x = ) ^ \y ( x ) xdx + J^^^*^ (x ) xdx

+ • • •+ ; -, l l ; ( x ) d x , m-1 T

45

"'- k+1 li=n '^ r

s imularly

-i C;i;'(x)dx = -i "I" 'il'^^xh' (x)dx , •m , ., ,. , ^-1 rk+1

k=n

and /m m-1 ,1 +1

- )nDG^'(x)dx = I )^ Zx^2•^|J<U)ix. m-1 /Ir+I

= 2 k r^^ T//'(x)dx ,

k=n ^ ^

by ignoring the discontinuity in the Riemann-Stieltjes

integral at x = k+1. Substituting these results in, we

have frn m—1 /V+"l

(5) )^\}J'M (x-CxH -4)dx = 2 )^ i/;'(x) xdx -I 2 C^^i/;'(x)dx

k=n - ^ m-1 "' • (k+1 , ., ,,

- E ^ )i T^'(x)dx. k=n

If v:e integrate the first integral on the right of (5)

by parts, we get

m-1 f]£+l i ^ - l , ; 2 K it/'(x) xdx = 2 i/;(ic+i) • ( -1) - '^WM k=n ^ ^ k=n^

^^^1 (k+1 I f , ^

k=n ^ ^

Substituting this into (5) and integrating the second and

46

third terms of (5), we have

(m I I m-1 )nV^ (x) ( x - H - i )c ix = 2 C(1^+1);(;(k+1) -k i / ; (k ) J

k=n ^ ^

m-1 - 2 I

k

^ /k+1 m.-l > ). iiy(x)dx - * £ C Ti;(k+l) --d/Ck)] =n ' " !s:=n '

m-1

K.—n

2 i//(k+i) - 2 i r v ^ dx k=n k=n

m - i 2 cv^(k+i) -w/(k):i,

k=n ^ ^

m-1 2 i / / (k+i)

k=n

m-1 m-1 i 2 l / / (k+l) + 1 2 ^ ( i ^ )

m-1

k=n

(k+1

k=n

\ ? . n V ( ) ' k=n

m-1 m-1 /Q = J: 2 \Lii'^+i) + i 2 i i ;(k) - L'u!/(x)dx,

k=n ^ k=n ^ ^ ^

m m-l i 2 \^(^) + - ^2"il;( ) - r^//(x)dx,

k=n+l K=n

1

k=n

- i \i;(x)dx, n -

m (m = 2 V^^^) " 4V^(n) - Sl//(m) - )^ l / ; (x)dx,

k=n Hence we have proved the lemma.

We return to the proof of one continuation of L (s). In the above lemma we put \hix) = x~s x"or s 9« 1 and in­vest Israte the remainder of the series for the zeta function.

47

E x-^ = *n^ + im^ + \l x-^ax - s "^ x-"-\x-CxD-*)dx. k=n ^ n ' -

.Now investigating the last term on the right, we see

0 < X -CxH < 1 for all x, orfx - C^J - ||< 1,

so it follows that

I s 'll x-^-1 ix - CxH - J)|dx <lsl i 'lx- -l I I x - H -ildx. n (ni _a-l

~ n

<'f a/n" - l/m°).

•" a or

sjl X-"-l (X - Cx: - 4)dX = 0 ( i | . ^f .

Hence if a > 0, the second integral (6) is bounded and

converges absolutely and uniformly as m ->oo. We now let

in ->oo in (6),

m / li2i 2 | s = l ini l /2n^+l /2m^+ ) ^ x"^dx m->a>k=n m->a>^ "

- s )^ x"^" (x -Cx] - i )dx y ,

(?) ^ 2 ^ i s = ^ + )n ^ ^ ^ ^ - s )^ x - " - \ x - C x > * ) d x ,

= 2E^ - \ l i - - S )^ X"^"-^ (x-[IxI]-^r)dX.

CD Now L (s) = 2 ^s ^^^ "'® w r i t t e n ,

^ k=l-^

(8) ^ ( s ) = " 2 i s + 2^^ - ^ 3 - - s C x - ^ " ^ x - C x > ^ ) d x ,

48

This is true for any n. If n=l, then (8) becomes

(9) H s ) = - -1- - s f^ x-^-1 (x- X -*)dx. *-* i-s -

The integral in (9) represents an analytic function for

o > 0, and hence t (s) is analytic for a > 0 except for

the' simple pole at s=l where the residue is 1. Hence we

have proved Theorem 4,6.

Theorem 4,7, If

f(s) = § ainl n=l nS

and g(s) = 1 £%I rfrl n^

and the series converge forOa^, and if f(s) = g(s) for

a > a^, then a(n) = b(n) , n=l,2,3.... .

Proof. Let

GO , .

do) F(s) = 2 ^ T ^ = 0 forvQ > OQ n=l n^

then c(n) = 0 for all n. Indeed, by Theorem 4.1, there

exists a region enclosing the real axis for s > a , in which

the series (10) converges uniformly. To prove that c(n) = 0

for all n, we suppose that c(m) is the first non-zero coef­

ficient. Then

(11) 0 = F(s ) = c(m) m -s i J c l m + l i f m + l ] - s U c ( m + 2 i [ m + 2 H . , 1 L V c(m) \ m J J c(m) V m / J ,

= c(m)m"'^ ( l + G(s)) ,

49

where G(s) = S (?- ) (E^] k=1 C(m) V m y

-s

If aQ< a3L< a, then

X r . - ( C - C T ) -ai

m+kj - < m+1j (m+k]

and

JG(S)| < 1 ( m+l\" °'" l m°l^ [c(m+k)l - lc(m)l \ m I 1 1 (m+k)oi .

which tends to 0 when s ->00 through real values of s.

Hence

11 + G(s)l > -I

for sufficiently large s; and (11) implies c(m) = 0, a

contradiction. It follows that if f(s) = g(s) for a > a^,

then a(n) = b(n) for all n. V/e shall occasionally refer

to this theorem as the identity theorem.

Theorem 4.8. If

00

^ ' = nil ^

CO ^ / V

and g(s) = 2 ^

converge absolutely for a > OQ , then for o > o^ ,

f(s)g(s) = I ai&i . i ^ n=l ^ n=l n

00 / N ^ c(n)

— ^.^ y^S n=l ^

where XtAAS TECHNOLOGlCAt COUtUfe LUBBOCK. TEXAS ^ LIBRARY

50

C(n) = 2 a(j) b(k) = 2 a(k) b(§) jk=n k m

= 2 a(§) b(k), kfn ^

the summation in the first expression being over all terms

a(j) b(k) for which jk = n and in the second and third ex­

pressions over all factors k of n.

Proof, Since for a > a^, the series f(s) and g(s)

are absolutely convergent, we may multiply them together

and rearrange terms.

CO . . 00 ^,, .

f(s) g(s) = 2 ^ ^ • 2 ^

j=l J k=l ^

When we multiply these two series together in the sense of

forming all possible products with one factor selected from

each series, the general term resulting is a(.l) b(k) _ si( O b(k)

(jk)s - 'i^

where n = jk. If now we add together all terms for which

n has a given value, we obtain a term c(n) n~^, where

c(n) = 2 a(j) b(k) , but jk=n

there is a term of the form c(n) n"^ for each value of n

as n ranges from 1 to oo .

Thus

f(s) g(s) = 2 c(n) n-s n=l

as required.

PART I I I

DIRICHLET'S THEOREM AND SOME

OF ITS CONSEQUExNfCES,

I'

,• '• n

CHAPTER V

PROOF OF DIRICHLET'S. TiiSOREM

V/e now prove Dirichlet's theorem which states that in

any arithmetic progression kn+a, n = 0, 1, 2, .,., where k

and a are relatively prime, there is an infinity of primes.

We let X be a character defined on the set of reduced

residue classes modulo k. Recall in Chapter III we devel­

oped the theory of characters for an Abelian group of order

h, hence it is necessary for us to show that the set of

reduced residue classes modulo k form an Abelian group.

Let S be the set of reduced residue classes modulo k,

together with the binary operation multiplication modulo k.

(1) S is a closed set under this operation; i.e. let s^

and Sp^S, then s-, Sp 6S; because, if we let s- = s-j (mod k) I » t

and So = Sp (mod k) where 0 < s-j_ < k and 0 < S2 < k.

Then s- Sp = s- S2 (mod k) , and since (s-j_,k) = 1 and

(sp,k) = 1 it follows that (s]_S2,k) = 1, hence 3 is closed, t

(2) Our operation is associative; i.e., let SQ_ = s-^ (mod k) Sp = Sp (mod k), and s^ = s^ (mod k) S, then

' ' * \ \

s-j_Sp = S-, S2 i'^^orl k) and S2S^ = S2 s^ (mod k)

^SiS2)s3 = s^ S2 s^ (i.od \) s-j_(s2S ) = 3- S2 s^ (mod k).

Hence (s-__"' )s3 = s^is^^s^) {•::io^ k). (3) There is an identity element e =i('-nod 'Oc 3.

52

53

t Let a = a (mod k)€S, then

ae = a (nod k),

ea = a (mod k),

(4) For every element a = a* (mod k) S, there is an in­

verse element which may be found by solving the congruence

ax = 1 (mod k).

(5) Our operation is commutative. This follows from the

commutativity of ordinary multiplication and the definition

of congruence modulo a fixed number. Hence we have shown

that S the set of reduced residue classes modulo k with our

binary operation is an Abelian group. The order of this

group is Cp(k), where <i)(k) is Euler's function.

V/e define the Dirichlet L-series as

00 . . L(s,X) = 2-^^-^ where s = a + it.

Theorem 5.1. If X 7 X-, the principal character,

then L(s, X) converges for a > 0.

Proof. In order to show th .t L(s, X) converges, we

need to find the abscissa of convergence a. By Theorem 4.5,

v:e can do this by determining the magnitude of n

S(n) = 2 Xi\n). We wish to divide the numbers m=l

m = 1 to n into coaplete residue systems mciulo k. To do

this we let n = qk + r where q is an integer and r is an

integer snrh that 0 < r < k. Then we have

S(n) = 2 (m) , m=l

54

X 2k 2 ^M + 2 ^M + m=l m=k+l

n=qk+r ••• + 2 ^(°^) •

m=qk+l

Now we wish to divide all except the last of these sums into

two parts, for example the first sum,

2 X(m) = 2 X(m) + 2 (in). m=l m=l m=l

(m,k)=l (m,k)>l

The first sum on the right extends over a reduced residue

system modulo k, and thus by Theorem 3.6. its sum i§ zero.

The second sum on the right is also zero because if (m, k)

> 1, then X(m) = 0 by definition of our character. In a

similar manner, we can show that the next q-1 sums are

each zero, so n=qk+r

S(n) = 2 ^(^). m=qk+l

and n=qk+r

S(n) < 2 IX(m)l, m=qk+l

By Theorem 3.3. X(m), m being an element of the reduced

residue class modulo k, is an ( (k) root of unity, and

hence X(m) can be written as

0 1 P £ Si'^) - 1.

Thus

2

X(m) = expL^I^^ • P

X(m) = 1 if m is an element of the reduced residue

or X(m)

system,

2 = 0 if m is not an element of the reduced

residue system,

55

So n=qk+r

S(n) < 2 1 < r . m=qk+l (m,k)=l

Thus S(n) = 0(1) = O(n^) for every 3 > 0, and by Theorem

4.5. a < 0 making L(s,x) converge fora > 0.

Corollary. The abscissa of convergence is actually

th 0 since for a< 0, the n term does not tend to zero.

Theorem 5*2. For a > 1

L(s,x) =Tr (1 - ^^i|i)-^ . •rx p S P ^

Proof. By definition we have

L(s.X) = 2 ^ n=l

This series is absolutely convergent since by the compari­

son test

»X(n)l ^ 1

and the series 2 | - ^ is convergent for a > 1. hence the

L(s,X) series is absolutely convergent. By Theorem 2.2.

we have X(n)v-1 S? X(n) _ TT fi .JLiili)

n=±

the desired result.

Theorem 5.3. L(s,/0 for the principal character X-j_

is analytic for a > 0 except for a simole pole at s = 1.

The residue at s = 1 is

5^

TT (i-i/p) = .dUii = h Plk ^ ^ .

Proof, Examining the principal character of primes,

we see

X^(p) = 1 if (p,k) = 1,

= 0 if (p.k) > 1. i.e. if p k.

From xheorem 5.2., we have

L(s.X) = (1 , Xi(p))-1 ^ p pS

= TT (1 . Xi(p)^~l TT ^ Xi(p))-1 (p.k)=l ps p|k P^

= TT (1 - ^ ) " ^ . (p.k)=l P^

But since

IT (1 - 1/p^)"^ = T[ (1 - i/ps)-l n (1 . l/pS)-l. P (p.k)=l p|k

we have

n (1 - i/pS)-i = n (1 - i/p^)"-^ n (i - i/pS) . (p,k)=l p p|k

so L(s,X) = n (1 - l/p^)~^ V (1 - /P^^ ' P v\^

= ^ (s ) • g (s ) ,

where g(s) = Ti (1 - 1/p^). Since g(s) contains only plk

finitely many terms and each of these terms is analytic,

the product is analytic. By Theorem 4.6 L (s) is analytic

except for a simple pole at s = 1; hence the L-se -ies is

analytic except for a simple pole at s = 1.

51

Residue at s = 1 : lim (s-1) lis) g(s) s->l ^

= lim (s-l)[i - -.1^ - si^ (x-[lxl|-fr)dx s->l

n (1-1/pS) Plk

lim s->l

[ ^ + 1 -s(s-l) i^ (x-dxH-Mx)^

TT (l-l/ps) ] . p|k

= n (1-1/p) , pjk

= Cfe(k)

_ h

We will now prove Dirichlet's theorem by a method

simalar to that used in Euler's proof in Chapter II.

L(s,X) is a complex valued function and has a multiple

valued logarithm. We will choose the branch of log L(s,X)

which is real when s is real and a > 1. For a > 1 we have

log L(s,X) = logTT (1 - ^ - i ) ' ^ P ^

= - 2 log (1 -p

X(p) TDS

)

For a > 1 x\re have X(p) TD'

1 < 1 so

los (1 - ) = - 1 ! ^ n=l ^P

00 (6) log L(s.X) = 2 E § #

p n=l ^

58

^ ps n=2 ^P ^ )

= 2 ^ - ^ E X(P^) p n=:2 ^P

• = 2 % ^ + R(s) . p P

Now we w i s h t o show R ( s ) i s bounded,

R(s)|< 2 P

00 X ( p )

± E 2

n=2 ^P

00

sns

no p n=2nP

00 , 1 ? E 2 rna

p n=2P

00

But 2 - ^ nto P^^

= d - p ' ^ ) " - * - f o r ( a > 0 ) , hence

CO 1 P - 2 ^

„ t 2 F ^ - i-p-° so

-2a lR(s)l l^2f9 0

< 4 . 1 1-2-0

-2a

P

00

2 n -2a 1-2-^ n= l

S i n c e ^ (2 )

a s s - > l ' ^ 0 .

CO = 2 n"^^ i s bounded a s s -> l ' ^^ , R ( s ) r e m a i n s

m=l

59

We now wish to isolate the first sum on the right-hand

side of (6) into two parts, nam.ely p = a (mod k) and p = a

(mod k). Before continuing directly with the general case,

we will look at the case a = 1. V/e now sum (6) over all

characters mod k, and by Theorem 3.4 there are cp(k) = h

of these characters.

Slog L(s.X) =£ 2 ^ +S R(s) . X X p P^ X

The series 2 ps is absolutely convergent since P ^

E p

X(p) 1 E h

which converges for a > 1. Since X is finite and 2 5-

is absolutely convergent, we can rearrange the order of

the terms in the series and obtain

2 log L(s,X) =2 2 ^ + 2 H(s) , X p X P X

2 - L 2 x(p) = 2 H(s) , p P X X

^ - ^ 2 x ( p ) p=l(mod k) X

2 - ^ 2 x ( p ) + 2 H(s). p^Kmod k) P X X

By Theorem 3.7, 2 X(p) can be expressed as X

60

^ h i f p = Kmod k) E X(p) = X 0 otherwTise

Thus

2 log L(s.x) = h 2 -4^+2 R(s) . X p=l(mod k) X

For the more general .case, we find a* such that

aa* = 1 (mod k) .

Multiplying both sides of (6) by X(a*) and summing over

all characters X modulo k, we have E X(a*) log L(s,X) =2X(a*) 2 ^ + 2 X(a*)R(s). X X p ^ X

Since X is finite and 2 ^ E^ is absolutely convergent P P"

2 X(a*) log L(s,X) = 2 - ^ 2 X(a*) X(p) + R*(s) X P ^ X

= 2 - ^ 2 X(a*p) + R*(s) p ^ X

2 ^ ^ 2x(a-p) a «-p=l(mod k) P X

+ 2 1 2 X(a*p)+R*(s), a*PEl(inod k) P ~" X

But by Theorem 3.7,

h if a'--p = Kmod k) 2 X(a- -p) = X 0 otherwise

Also observe that if a*'p=l(mod k). then p = aa^p

=a(mod k), so

61

(7) 2 X(a*) log L(s.X) = h 2 p=a(mod k) ps

+ R*(s),

Now looking at R*(s) = 2 X(a*) R(s), X

we see

1R*(S)1< 21 X(a*) R(s) X

< 2 |H(S)

X

and since R(s) is bound by Theorem 3,7."

R*(s)|< h.

Thus R*(s) remains bounded as s->l'*" , L(s,X) for X 4 Xj^

is analytic at s = 1 by Theorem 5.I. and remains bounded as

s->l**" ; thus log L(s,X) for X 5 X- rem.ains bounded unless

L(1,X) = 0. Also by Theorem 5,3. we see as s->l'*" , L(s,X-j_)

->00 . Hence 2

1 -1+0 . , .

•^ ->oo as s->l , provided p=a(mod k)

L(1,X) 5 0 for X 5 X-L .

Theorem 5.4. If X 5 X^, then L(s,X) 4 0.

Proof. From (6) we have

(8) logTT L(s,X) = 2 l o s L(s,X) . X X

. ^ S f X(pn) X p n=l ^P

CO

n=l p "P X

62

2 ^ 2 , ; ^ 2 x ( p ^ ) n = l p^^sKmod k) ^P X

CO ^

\ i l p lfmod k)^^F^ f ^^P''^' 00

= h 2 2 ^ n ^ l p nP^^

by Theorem 3 . 7 , Now l e t

h I 2 - ^ = 2 ^ = f(s) , n = l p np^^ m=l ^^

where a(m) d e n o t e s t h e c o e f f i c i e n t s of t h e D i r i c h l e t s e r i e s

and f ( s ) i s t h e sum of t h i s s e r i e s . Note t h a t 1 > a(m)

> 0 . Now l e t a be t h e a b s c i s s a of c o n v e r g e n c e , t h e n

F L ( S , X ) = e^(^) = 1 + f(s) + I ^ + . . . I tAsl X 2 n=0 ^ - .

• C D , X

Now f ( s ) = 2 ^ ? and so f o r a > cc, m=l ^

f ^ ( £ ) ^ ^ a ( - ^ ) ( k )

^ - k = l 1 ^

where we have t a k e n t h e p r o d u c t of n D i r i c h l e t s e r i e s of

t h e form 2 ^ ^ t)y Theorem 4 . 8 . Moreover s i n c e 1 > mti ^^

(n) a(m) > 0 i t f o l l o w s t h a t a ^ ^ ( k ) > 0 . Thus by (8) and

( 9 ) . i f a > a 00

(10) TT L(s,X) = 1 + 2 2 § a(^)(k) m=l k = l ' ^

CO T CQ (rr)

1 -f 2 1 2 a^^^^k) , k = l ^ m=l

63

.g b(k) k"s"

the interchange of summation being justified by absolutely

convergent series (the terms are all non-negative).

The idea of the proof that L(1,X) 4 Q, is as follows.

The left-hand side of (10) has a simple pole at s = 1

arising from L(s,X^), the other L series being regular at

S = 1. If L(1,X) = 0, then this zero would cancel the

pole making the left-hand side regular. But we shall show

that the ri>2-ht-hand side has a singularity in the interval

0 < s < 1. This is a contradiction."

From (9) for a > 1, CO S" a m ) < ^ a(m)^ ^ 1 m=l m=l m=l

which converges for a > 1. Thus f(s) converges for a > 1,

This means a< 1, V/e now wish to show that 1/h < a by

showing f(s) diverges at 1/h where h = (i>(k).

f(l/h) = h f 2 l/m P*"" -2 , 2 -THp m=l p° =l(mod p) m=jh p^=l(mod p)

Now by the Euler-Fermat theorem, if pfk, then p sp* =

Kmod k) C15I]. Hence if we sum over all primes such that

p = l(mod p), x\re sum over all primes such that p f k and

n all primes p which divide k and p = 1 (mod k) so

f (1/h) > h ^ E THHi- > h 2 2 -k^

64

00

= h2 E-^= E E P+k m=jh JhpJ p+ic j = i pJ

> E Y- E ^' E I . ptk P p P pfk P

3y Theorem 2,3, E 1/P diverges, and the second sum has P

only finitely many terms. Thus the series f(l/h) diverges

and a > 1/h, CO

By Theorem 4.4., f(s) = E ^^%^ has a singularity at

m=l °^

s = a since a(m) > 0. Let 3 be the abscissa of convergence

of the Dirichlet series for exp (f(s)). If exp (f(s)) con­

verges, then f(s) must converge and vice versa, so a = p

where a is the abscissa of convergence for the Dirichlet of

f(s). Since the coefficients of exp (f(s)) are positive,

exp (f(s)) has a singularity at P = a. This m.eans that TT L(s,X) has a singularity at 3 and 1/h < 3 < 1. On the X

other hand, since L(s,X) for X 5 X-j is regular for a > 0,

it follows that TT L(s,X) is regular for 0 < o < 1 except

X

possibly for a simple pole at s = 1, since L(s,X-|_) is regu­

lar except for a simnle pole at s = 1.

If now L(1,X) = 0 for some X (say X = X2) then

lim L(s,X2) = A s->oo s-1

some f i n i t e c o n s t a n t . T h e r e f o r e , u s i n g Theorem 5 . 3 ,

l i m TF L ( s , X ) = l i m ( s - 1 ) L(s,X-,) ^ ^ ^ ' f ^ ) TT L( s .X) , s->oo X s->ao ^'^ l-Ai 2

Xy^x;

e5

= § . A TT L(s,X) = finite limit.

X^X2

That is, TT L(s,X) is regular for 0 < a < 1. This contra­

dicts the fact that it has a singularity at 3.

The proof of the nonvanishing of L(1,X) and hence of

Dirichlec's theorem is complete.

CHAPTER VI

SOME CONSEQUENCES OF DIRICHLET'S

• THEOREM

Dirichlet's theorem is related to the general theory

of the "qualitatives" distribution of prime ideals. Its

significance is seen in the following way.

Let A. be any set of prime numbers. V/e wish to con­

struct a measure of that proportion of all primes which are

contained in the set A. One such measure can be defined

as follows:

s->1^0 P6A

2 1/P^ = ci. P

The upper sum is taken over all Drim.is in the set A, while

the lower sum is taken over all primes. This lim.it is

called the Dirichlet density and is abbreviated as D,D.

We note that the existence of a positive Dirichlet

density for a set A indicates that the set contains infi­

nitely many primes. If for d > 0 there vjere only finitely

many primes in A, then as s -> 1" ^ the nur.era.3r would re-

m.a n finite as the denominator ->oo ; the ratio of the

numerator to denominator ', ould -> 0 contradicT:inr- the fact

that d > 0.

66

61

Theorem 6 . 1 . If ( a , k ) = 1 . t hen

lim^ 2 - ^ s~>K^p=a(mod k) P ^ 1

-nS P P

The primes are thus "equidistributed" in all reduced resi­

due classes modulo k.

Proof. From equation (8) in Chapter V we have,

2 X(a*) log L(s,X) = h 2 4 - + R*(s) . X p=a(mod k) P

or

2 log L(s.X) + 2 X(a*) log L(s,X) = h 2 ' "Ts + R*(s). X-L X?X-L p=a(mod k)P

where we recall R * ( s ) . and log L(s,X) for X 5 X^ remain

bounded as s->l .

Also in the proof of Theorem 5.3. we saw

L(s,X. ) = tis) TT (1 - i ) . ^ ^ plk P

Now taking logarithms we have

log L(S,XT ) = log r (s) + log TT (1 - is) . ^ ^ plk ^

but

log fis) = - 2 log (1 - P'^) » P

00 ^

= 2 EA p k=l kP*^

68

= 2^^2 P^ p k=2'"kP^ .

= 2 - ^ + R(s) , p P^

where R(s) remains bounded as s-> l"*" .

So

log L(s.Xn ) = log i is) + log TT (1 - -1-.)

p l k p ^

= 2 - ^ + R(s) + log n (1 -p P plk

and

log L(S,X-L) + 2 X(a*) log L(s,X) = h X5^X

+ R- (s)

p=a(mod k) P'

becomes

2^3 +^(s) + log TT (1 - 1 ) + 2 X(a*) log L(s,X) plk

: h

X? X 1 1

p=aTmod k) P + R-(s).

We now divide each term by 2 -^^ and take the limit as

+0

p

1 +0 s->l' . We recall that 2 — ^ ~ > ^ ^s s->l and also

P ^

2 p=a(mod k) ^

1 -. +0 , -, +0 , 4. . 4. •^ ->00 as s->l . As s->l eacn quotient

+0 -> 0 as s->l except for the quotient

h E psa(modk)

2 ^ p

^ ~ ^

which approaches 1, thus

as s -> 1

p=a(mod k) P'

2 fs P ^

+0

- > 1 "^TkT

LIST OF REFERENCES

1. G.H. Hardy and E.M. V/right, The Theory of Numbers, Oxford University Press, Clarendon, I938, 245,

2. E. Hills, Analytic Function Theory, vol. I. Blaisdell, 1963. 108.

3. L.L, Small, Calculus. ApDleton-Century-Crofts. New York, 1949, 384-386.

4. T. Port, Infinite Series. Oxford University Press. Clarendon, 1930, 26.

5. L.L. Small, Elements Of The Theory of Infinite Processes, McGraw-Hill. New York. 1923. 224.

6. T.J, Bromwick, An Introduction To The Theory Of Infinite Series, Macmillan and Company, London, I926, 107-108,

7. L. L. Smail, Calculus. Appleton-Century-Crofts, New York, 1949, 378.

8. I.I. Hirschman, Jr., Infinite Series. Holt, Rinehart and Winston, New York, I962. 28.

9. L.L. Smail, Calculus, Appleton-Gentury-Crofts, New York, 1949. 381.

10. M. Hall, Theory of Group's. Macmillian Company, New York. 1959. 285-291.

11. E. Hille. Analytic Function Theory, vol. I, Blaisdel, 1963, 192.

12. E. Hille. Analytic Function Theory, vol. I. Blaisdel. 1963. 126.

13. S. Hille, Analytic Function Theory, vol. I. Blaisdel. 1963, 116.

14. W.J. Le Veque, Topics In Number Theory, vol. I, Addison Wesley, Reading, 1956, 42.

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