Download On the Number of Prime Numbers less than a Given Quantity

On the Number of Prime Numbers
less than a Given Quantity
(MATH 618 project - May 2, 2000)
Tzanio V. Kolev
1
Introduction
“It will be another million years, at least, before we understand the primes.”
P. Erdös
Prime numbers are probably one of the most beautiful objects in all of mathematics. It
is remarkable, that they have such a simple definition: “p is prime iff p has no other
divisors, besides 1 and p”, and at the same time their properties are so hard to explore.
The importance of the primes was realized in antiquity, when it was proved that in some
sense they are the “building blocks” of the set of the positive integers N. Nowadays, the
modern applications of prime numbers in areas such as physics, cryptography, and coding
theory make any information that can be obtained about them of great value.
In this paper we are interested in the distribution of prime numbers in M, where M ⊆ N
is a given infinite set. This is a fundamental question, and it has been analyzed by some
of the greatest mathematicians of all time, such as Euclid, Euler, Gauss, Dirichlet, Chebyshev,
Riemann, Landau, Wiener, Hardy, Erdös, and many more. Their efforts gave birth to many new
important mathematical ideas, e.g. analytic number theory, Riemann zeta function, and
tauberian theorems.
To make the problem concrete, let pn denote the nth prime in M, π(x) be the number of
the primes in M not exceeding x, and δn = pn+1 − pn . Here are some general “distribution”
questions:
(a) Are there infinitely many primes in M
(b) Is there a useful formula for pn or π(x)
(c) Is there an asymptotic formula for pn (as n → ∞) or π(x) (as x → ∞)
(d) How big the gap between two successive primes can be (lim supn→∞ δn )
(e) What can be said about lim inf n→∞ δn
(f) What is the probability for a randomly picked number to be a prime
In the simplest case M ≡ N, (a) has a positive answer and it was proved by Euclid using one
of the first examples of “prove by contradiction” technique. However, if M 6≡ N this can
be a very difficult question. A variety of formulas for pn and π(n) exist, but all of them
are contrived to such an extent that they are of little practical value. For example, if [x]
1
denotes the integer part of x, i.e. the unique integer such that x − 1 < [x] ≤ x, then for
n ≥ 3 (see [15]):
π(n) = −1 +
n X
j=3
(j − 2)!
(j − 2)! − j
j
n
and
pn = 1 +
2
X
f (n, π(j)),
j=1
where f (x, y) = (1 + (x − y)/|x − y|)/2 if x 6= y and 0 otherwise. Question (e) still
remains open, although the hypothesis that there are infinitely many twin primes, i.e. that
lim inf n→∞ δn = 2 has been well known for many centuries. One estimate connected to
this problem is lim inf n→∞ δn ≤ (1/4 + π/16) log pn , (see [16]). Another related result is a
theorem of Brun (see [4]) which states that the series of the reciprocals of all twin primes
converge. As for (d), it has an easy answer and it is that the gaps can be infinitely large,
i.e. lim supn→∞ δn = ∞. Indeed, for any n ∈ N it is clear that there is no prime in the set
{n! + 2, n! + 3, · · · , n! + n} so lim supn→∞ δn ≥ n. It turns out that (c) is the question which
can be answered in the greatest details. If we use the following definition of equivalence
f (x) ∼ g(x) ⇐⇒
lim
x→∞
f (x)
= 1,
g(x)
then the asymptotic behavior of π(x) as x → ∞ is given by:
Theorem 1 (Prime Number Theorem)
π(x) ∼
x
.
log x
(1)
A straightforward corollary from this result is that pn ∼ n log n. Indeed n = π(pn ) ∼
pn / log pn , so log n ∼ log pn , therefore pn ∼ n log n. Another implication from (1) is that
the answer to (f) is 0, because limx→∞ π(x)/x = 0. In fact, the prime number theorem is
equivalent to saying that the probability for a randomly picked number less than x to be a
prime is 1/ log x.
In the next sections we will develop the machinery needed for proving Theorem 1. A short
outline of the history behind the problem is given in §2. In §3 we explore the Riemann
zeta function ζ(z) and the zeros of its analytical continuation. Finally, a proof of the prime
number theorem is given in §4.
The following notation will be used: P = {2, 3, 5, · · · , 26972593 − 1, · · ·} is the set of all prime
numbers, p stands for any element of P, e.g.
X
X
:=
,
pm <x
{p∈P : pm <x}
m|n means that m divides n i.e. m, n, n/m ∈ N, log(·) is the principal value of the logarithmic function (the branch which is real on the real axis). Also
f (x) = O (g(x))
⇐⇒
lim sup
f (x) = o (g(x ))
⇐⇒
x→∞
2
x→∞
lim
|f (x)|
< ∞,
g(x)
|f (x)|
= 0.
g(x)
2
Historical background
“Till now the mathematicians tried in vain to discover some order in the
sequence of the prime numbers and we have every reason to believe that there
is some mystery which the human mind shall never penetrate.”
L. Euler
Although we are not sure when the conception of a prime number was clearly formulated, we
know that in the time of Euclid of Alexandria (325BC-265BC) the following facts concerning
the primes were already known:
• Theorem 2 (Fundamental Theorem of Arithmetic) Every positive integer can
be represented as a product of primes. This representation is unique up to a rearrangement of the factors.
• Theorem 3 (Euclid) There are infinitely many primes.
• The primes can be effectively listed using the sieve of Eratosthenes algorithm.
Probably, the actual investigation of the distribution of the primes began with the work of
Leonhard Euler (1707-1783), who proved Theorem 2 using the following argument:
Assume that {p1 , · · · , pN } is the complete list of all primes, and consider the product
N Y
k=1
1
1−
pk
−1
=
N Y
k=1
1
1
1+
+ 2 + ···
pk
pk
, since
1
< 1.
pk
The right-hand side is a finite product of absolutely converging series, so
N Y
k=1
1
1−
pk
−1
=
∞
X
k1 =0
···
∞
X
kN =0
1
pk11 · · · pkNN
≥
R
X
1
n
n=1
∀R ∈ N , (by Theorem 2).
However, the series on the right diverges as R → ∞. This is a contradiction.
Q
Let us go one step further, and observe that since 0 < 1/p < 1 the product p (1 − p−1 )−1
P
P
−1
converges if and only if
p log (1 − p ) converges that is if and only if
p 1/p converges.
Now, the Euler’s proof in fact shows that
Y
X1
1 −1
1−
= ∞ =⇒
= ∞ =⇒ there are infinitely many primes.
(2)
p
p
p
p
The first published statement (see [19]) concerning π(x) is due to Adrien-Mari Legendre (17521833). He asserted in 1798 that π(x) ∼ x/(A log x + B) for some constants A and B. Later
he refined his conjecture to be
π(x) =
x
log x + A(x)
, where A(x) is “approximately 1.08366”.
(3)
As we already know this is almost correct. In fact the approximation (1) can be obtained
just by setting A(x) ≡ 0 and then taking the limit x → ∞.
Around the same time (1792-1793), Johann Carl Friedrich Gauss (1777-1855) had already done
extensive work on the theory of primes. He considered the tabulation of prime numbers as
3
some sort of pastime and compiled tables of their distribution in various intervals of length
1000 (see [10]). Gauss suspected that the density with which primes occur in a neighborhood
of n is 1/ log n, so the number of primes in the interval [a, b) should be approximately
Z x
Z b
dt
dx
=⇒ π(x) ∼ Li(x) :=
.
(4)
2 log t
a log x
It is not difficult to check that both approximation (1) and (4) give the same asymptotic.
Indeed, by l’Hôpital’s rule:
Li′ (x)
log x
Li(x)
= lim
= 1.
= lim
2
x→∞ (log x − 1)/ log x
x→∞ log x − 1
x→∞ x/ log x
lim
Nevertheless, the approximation (4) is better than (1). For example if n = 106 , then
n
π(n) = 78498, [Li(n) − π(n)] = 128,
π(n) −
= 6115.
log n
Moreover, it can be proved that there exists a positive constant a such that
√
|π(x) − Li(x)| = O(xe−a
log x
) but
√
lim sup |π(x) − x/ log x| / (xe−a
x→∞
log x
) = ∞.
The first result concerning prime numbers in M 6≡ N was given by Johann Peter Gustav Lejeune
Dirichlet (1805-1859), who proved in 1837 (see [7]) a problem stated by Legendre in [20]:
Theorem 4 (Dirichlet) In any arithmetic progression {d n + b}n∈N with first term coprime to the difference there are infinitely many primes.
The idea of the proof was to generalize Euler’s method i.e. to use that
X
1
= ∞ ⇐⇒ there are infinitely many primes in {d n + b}n∈N .
p
p≡d(mod q)
Dirichlet’s work introduced a fertile new idea into the number theory: the use of the analytic
methods and was the beginning of a revolution in number-theoretic thought.
A major progress in estimating π(x) was due to Pafnuty Lvovich Chebyshev (1821-1894). Still
using methods of an elementary combination nature, he proved in 1850 the following conjecture formulated in 1845 by Joseph Louis Francois Bertrand (1822-1900):
Theorem 5 (Chebyshev) For n > 3 there is at least one prime between n and 2n − 2.
He also proved the estimates (see [6, 5]):
π(x)
9
π(x)
7
≤ lim inf
≤ 1 ≤ lim sup
≤ ,
x→∞ x/ log x
8
8
x→∞ x/ log x
(5)
which imply that if limx→∞ x/π(x)
log x exists its value should be 1. However, Chebyshev was
unable to prove that this limit exists.
The first big step toward a proof of the prime number theorem was made by Georg Friedrich
Bernhard Riemann (1826-1866). In his memoir [29] he connected arithmetic with the theory
of functions of a complex variable by introducing
ζ(z) =
∞
X
1
,
nz
n=1
4
(6)
which became known as the Riemann zeta function. Of course this function was already
considered by Euler, but it was named after Riemann, because he was the first to explore it
for complex values of z. Clearly the series on the right converges absolutely and uniformly
on compact subsets of the half-plane ℜz > 1, and therefore ζ(z) is well defined and analytic
for ℜz > 1. Moreover, repeating the same argument as in the Euler’s proof of Theorem 2
one can obtain that if ℜz > 1 then
Y
1 −1
.
(7)
ζ(z) =
1− z
p
p∈P
Q
Indeed, let ζX (z) = p∈PX (1 − p−z )−1 , where PX = {p ∈ P : p ≤ X} = {p1 , · · · , pN }
are all the primes up to X. Because of the absolute convergence one can change the order
of summation, i.e.
ζX (z) =
∞
X
k1 =0
···
∞
X
kN =0
1
pk11 · · · pkNN
z
=
X 1
+ R(z).
nz
n≤X
Here R(z) is the sum corresponding to all numbers n > X whose prime divisors are all
in PX . Clearly if z = σ + it then
|R(z)| ≤
X 1
→ 0 as X → ∞
nσ
, since σ > 1.
n>X
Thus ζX (z) → ζ(z), as X → ∞.
Riemann proved that ζ(z) can be continued to a meromorphic function on C with a pole in 1,
with residue 1. The Euler’s proof shows that the presence of the pole at 1 implies that there
are infinitely many primes. However, the connection between ζ(z) and the distribution of
primes runs even deeper. Riemann used the Möbius function (1832):

if n = 1
 1
m
(−1)
if n is a product of exactly m distinct primes ,
µ(n) =
(8)

0
otherwise
to define the so called Riemann prime number formula:
R(x) =
∞
X
µ(n)
n=1
n
√
Li ( n x ),
(9)
which is even better approximation to π(x) than (4), e.g. |π(106 ) − R(106 )| = 29. It turns
out that besides the trivial zeros of (the analytically continued) ζ(z) in {−2, −4, −6, · · ·} all
other zeros lie in the strip 0 < ℜz < 1 called the critical strip. There is a direct connection
between this non-trivial zeros {ρ} and π(x):
X
π(x) = R(x) −
R(xρ ).
ρ
Also, the prime number theorem is equivalent to the fact that there are no non-trivial zeros
on the boundary of the strip. The famous Riemann hypothesis asserts that all non-trivial
zeros lie on the critical line ℜz = 21 . This conjecture is one of the most celebrated open
problems in all of mathematics.
5
The first proofs of the prime number theorem were given independently in 1896 by Jacques
Salomon Hadamar (1865-1963) and Charles Jean Gustave Nicolas de la Vallée Poussin (1866-1962)
(in [13, 28]). Although they differ in details both proofs follow Riemann ideas and establish
the existence of a zero-free region for ζ(z) in the critical strip. Precisely, the following
asymptotic was proved:
√
(10)
π(x) = Li(x) + O(x e−a log x )
where a is some positive constant. In the same paper de la Vallée Poussin proved also the
following generalization of the prime number theorem: if π(x; d, b) is the number of primes
not exceeding x in the arithmetic progression {dn + b}n∈N where (d, b) = 1, and φ(d) :=
#{m ∈ N : m < d, (m, d) = 1}, then
π(x; d, b) ∼
1
x
.
φ(d) log x
(11)
The error term in (10) depends on what is known about the zero-free region, mentioned
above. As the knowledge of the size of this region increases, the error term decreases. For
example, in 1901 Niels Fabian Helge von Koch (1870-1924) showed that the Riemann hypothesis
is equivalent to
√
(12)
π(x) = Li(x) + O( x log x).
In 1903 Edmund Georg Hermann Landau (1877-1938) gave a proof of the prime number theorem
which contained a considerable amount of simplifications. Further simplifications were made
by Norbert Wiener (1894-1964) and his student Shikao Ikehara. They applied Fourier analysis
methods in order to obtain an important class of analytic results, known as tauberian
theorems (see [35]). These proofs again depended on the behavior of ζ(z) in C and were
indisputably analytic in nature.
It was natural to ask if one could prove the theorem by a method not involving complex
functions theory. By 1930 many mathematicians had the feeling that such a proof could
not exist. The reason was the assumption that any proof should use the equivalence of the
prime number theorem and ζ(z) being non-vanishing on the boundary of the critical strip.
However, these heuristic reasons were false. In 1949 Paul Erdösh (1913-1996) and Atle Selberg
(1917-) succeeded in giving several elementary proofs without using any analytic methods
(see [9, 31]). In all of them the starting point was the relationship
X
X
log2 p +
log p1 log p2 = 2x log x + O(x),
(13)
p≤x
p1 p2 ≤x
which is known as the Selberg’s inequality. Using a modified version of this result Selberg
proved the Dirichlet’s Theorem 4 in “elementary” way the same year (see [30]).
It has been checked and always found that π(n) < Li(n). However, Skewes proved (see
1034
[32, 33]) that the first crossing of π(n) = Li(n) occurs before 1010 . This number was
known as “the largest useful number in mathematics”. Since then, the upper bound for the
crossing point has subsequently been reduced to 10371 . John Edensor Littlewood (1885-1977)
proved in 1914 (see [23]) that the inequality π(n) < Li(n) reverses infinitely many times for
sufficiently large n.
Further information on the history of the prime number theorem and related topics can be
found in [11, 21] and online at [36, 37, 38, 39].
6
3
ζ(z)
“8. Problems of prime numbers ... to prove the correctness of an exceedingly
important statement of Riemann, that the zero points of the function ζ(z)
... all have the real part 1/2, except the well-known negative integral real zeros”
International Congress of Mathematicians, Paris, 1900
D. Hilbert
In this section, we prove some results for the Riemann zeta function defined for ℜz > 1
by (6). We have already obtained the equality (7), which clearly shows that there is a
connection between ζ(z) and the prime numbers. In order to explore this connection we
should first continue ζ(z) to a meromorphic function on C. Our approach will follow [17].
3.1
Analytic continuation in C. Functional equation
We begin with a theorem, which gives an integral representation of ζ(z) in ℜz > 1.
Theorem 6 If ℜz = σ > 1 then
Z ∞
z 1
−z/2
xz/2−1 + x−z/2−1/2 ω(x) dx,
ζ(z) =
+
π
Γ
2
z(z − 1)
1
P
−πn2 x .
where ω(x) = ∞
n=1 e
(14)
Proof: One of the definitions of Euler’s gamma function is the integral formula
Z ∞
Γ(z) =
e−t tz−1 dt , for ℜz > 0.
(15)
0
If σ > 0, then clearly
Γ
z 2
=
Z
∞
e−t tz/2−1 dt,
(16)
0
2
πn x
which after the change of variables t =
gives
Z ∞
Z ∞
z
z 2
2
Γ
e−πn x xz/2−1 dx. (17)
= nz π z/2
n−z =
e−πn x xz/2−1 dx =⇒ π −z/2 Γ
2
2
0
0
We sum this over all n ∈ N and assume that σ > 1, so we have ζ(z) on the left-hand side:
∞ Z ∞
z
X
2
−z/2
ζ(z) =
e−πn x xz/2−1 dx.
(18)
π
Γ
2
0
n=1
P
P∞
−πn2 x < (πx)−1
−2 converges,
Since ex > x for any fixed x > 0, the series ∞
n=1 e
n=1 n
so ω(x) is well defined in (0, ∞). Therefore, we can change the order of summation and
integration:
∞ Z
X
n=1 0
∞
[·] = lim
N →∞
N Z
X
∞
[·] = lim
Z
N →∞ 0
n=1 0
N
∞X
[·] =
n=1
Z
0
∞
∞X
[·] − lim
n=1
Z
N →∞ 0
∞
X
[·].
n>N
Thus the right hand side of (18) can be divided in two parts:
Z ∞X
Z ∞
∞ Z ∞
X
2
z/2−1
−πn2 x z/2−1
e−πn x xz/2−1 dx.
ω(x) x
dx − lim
e
x
dx =
n=1 0
N →∞ 0
0
7
n>N
(19)
111111111
000000000
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000111111111
111111111
000000000
000000000
111111111
000000000
000000000111111111
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
000000000
000000000111111111
111111111
000000000
111111111
000000000
111111111
000000000
000000000111111111
111111111
000000000
111111111
000000000111111111
111111111
000000000
000000000
111111111
0000000000
1111111111
000000000
000000000111111111
111111111
0000000000
1111111111
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
000000000111111111
111111111
0000000000
1111111111
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
111111111
000000000111111111
111111111
0000000000
1111111111
000000000
000000000
111111111
0000000000111111111
1111111111
000000000
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
111111111
000000000
111111111
000000000111111111
111111111
0000000000111111111
1111111111
000000000
000000000
111111111
0000000000000000000
1111111111
000000000111111111
111111111
000000000
000000000111111111
111111111
0000000000
1111111111
000000000000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
000000000
111111111
000000000
111111111
000000000111111111
111111111
0000000000111111111
1111111111
000000000111111111
111111111
000000000
000000000
111111111
0000000000
1111111111
000000000000000000
111111111
000000000
111111111
000000000
111111111
000000000
111111111
0000000000
1111111111
000000000
000000000
111111111
000000000
111111111
000000000111111111
111111111
0000000000111111111
1111111111
000000000
111111111
000000000
000000000
000000000
111111111
0000000000111111111
1111111111
000000000111111111
000000000
000000000
111111111
000000000111111111
111111111
0000000000
1111111111
N
Figure 1: Why
P∞
n=N +1 f (n)
N+1
≤
N+2
R∞
N
N+3
N+4
f (x)dx, for f (x) - positive and decreasing.
2
Our first task is to prove that the second integral tends to 0. Since e−πn x is a positive,
decreasing sequence the Riemann sums are smaller than the integral (see fig. 1), i.e.
Z ∞
Z ∞
X
1
−πt2 x
−πn2 x
√
e−y y −1/2 dy , (change y = πt2 x).
(20)
e
dt =
e
<
πx
2
2
πxN
N
n>N
In order to estimate this integral we need another property of Γ(z) namely that
z 6∈ Z =⇒ Γ(z)Γ(1 − z) =
π
.
sin πz
(21)
Applying (21) for z = 1/2 gives
Z ∞
√
1
Γ
e−t t−1/2 dt = π,
=
2
0
√
thus one upper bound for the integral in (20) is 1/(2 x). On the other hand
Z
∞
πxN 2
e−y y −1/2 dy ≤ √
1
πxN 2
so we can deduce that
X
n>N
−πn2 x
e
≤ min
(
2
∞
e−πxN
e−y dy = √
,
πxN 2
πxN 2
Z
2
1 e−πxN
√ ,
2 x 2πxN
)
.
Consequently
Z
Z
Z ∞ −πxN 2
∞X
1/N
e
1 σ/2−1
2
−πn x z/2−1
√ x
dx +
e
x
dx ≤
xσ/2−1 dx
0
2 x
2πxN
1/N
0
n>N
(σ−1)/2
Z
1
1 −σ/2 −σ+1 ∞ −u σ/2−2
1
+ π
N
e u
du , (u = πN 2 x)
=
σ−1 N
2
πN
(σ−1)/2
σ−1
1 −σ/2 −σ+1
1
1
−1/2
+ π
N
(πN )
Γ
≤
σ−1 N
2
2
= C1 N −(σ−1)/2 + C2 N −(σ−1/2) −→ 0 , when N → ∞.
8
(22)
This implies (see (18), (19)) that for σ > 1 we have the equality
Z ∞
z
−z/2
ω(x) xz/2−1 dx.
(23)
ζ(z) =
π
Γ
2
0
At this point we observe that for x > 0, ω(x) satisfies the following identity (see [17], p. 5-8):
√
x √
1
+ x ω(x).
ω(x−1 ) = − +
2
2
Let us apply this to the integral
Z ∞
Z 1
z/2−1
ω(x−1 ) x−z/2−1 dx
, (x 7→ x−1 )
ω(x) x
dx =
1
0
Z ∞
1
=
ω(x) x−z/2−1/2 dx.
+
z(z − 1)
1
Thus
Z ∞
ω(x) x
z/2−1
dx =
0
Z
1
+
0
Z
∞
1
1
=
+
z(z − 1)
Z
∞
1
ω(x) x−z/2−1/2 + xz/2−1 dx.
(24)
The theorem follows from (24) and (23).
Let I(z) be the improper integral on the right-hand side of (14). Our next step is to prove
that I(z) defines an entire function of z. Fix m ∈ R, K− a compact subset of ℜz > m and
let M = maxz∈K |z| < ∞. For z ∈ K and x > 1 we have
−z/2−1/2
+ xz/2−1 ω(x) ≤ x−m/2−1/2 + xM/2−1 Ce−πx , where C 6= C(x), (25)
x
because m < ℜz ≤ |z| ≤ M , and ω(x) = O(e−πx ) as x → ∞:
eπ(n
2 −1)x
> π(n2 − 1)x =⇒ eπn
2x
>
π πx 2
e n
2
, for n ≥ 2.
(26)
The estimate (25) implies, that the integral I(z) converges absolutely and uniformly on the
compact subsets of ℜz > m for all m ∈ R, i.e. that the sequence
Z n
(27)
xz/2−1 + x−z/2−1/2 ω(x) dx
In (z) =
1
converges to I(z) absolutely and uniformly. On the other hand the integrand of In (z) is
holomorphic, so we can differentiate under the sign of the integral
Z n
∂ z/2−1
∂
(28)
x
+ x−z/2−1/2 ω(x) dx = 0.
In =
∂ z̄
1 ∂ z̄
Thus I(z) is holomorphic on ℜz > m as an uniform limit of holomorphic functions for all
m ∈ R, i.e. I(z) is an entire function of z.
Remark 1 Another way of proving that I(z) is an entire function of z is by the Morera’s
theorem, noting that the absolute convergence allows us to switch the order of integration
(Fubini’s theorem).
We proved Theorem 6 under the assumption that ℜz > 1, but as we just have shown the
right-hand side is defined in C−{0, 1}.
9
Lemma 1 If f, g : U 7→ C ∪ {∞} are meromorphic on a domain U ⊂ C and if f ≡ g on
some nonempty open set M ⊂ U then f ≡ g on U .
Proof: By the Weierstrass’s theorem there exist holomorphic functions in U : fn , fd , gn , gd ,
such that (see [12], p. 271):
gn
fn
f≡ , g≡ .
fd
gd
Now, f ≡ g on M implies that the holomorphic functions fn gd and fd gn are equal in M
and therefore in U . Consequently f and g have the same (discrete) singular set, so
lim [f (z) − g(z)] = lim
z→z0
z→z0
fn (z)gd (z) − fd (z)gn (z)
= 0,
fd(z)gd (z)
for every z0 ∈ U . Thus f ≡ g on U .
Lemma 1 implies, that the formula (14) can be used to obtain the unique analytic continuation of ζ(z) in C. To find the singular points of this continuation recall that
Γ(z) has simple poles in −N0 := {0, −1, −2, · · ·},
z 6∈ −N0
(29)
(−1)k
,
k ∈ N0 =⇒ Res(Γ, −k) =
k!
"
#
∞ 1Y
1 z
z −1
=⇒ Γ(z) =
=⇒ Γ(z) 6= 0,
1+
1+
z
j
j
(30)
(31)
j=1
and observe that the simple pole of Γ(z) at 0 cancels with the simple pole of 1/z on the
right. Putting all this together produces:
Corollary 1 The function
f (z) = π
z/2
Γ
−1
z 2
1
+
z(z − 1)
Z
1
∞
z/2−1
x
+x
−z/2−1/2
ω(x) dx
(32)
is meromorphic in C with a simple pole in 1 with residue 1. It is equal to ζ(z) on the
half-plane ℜz > 1, and thus f (z) is the analytic continuation of ζ(z) in C.
Observe that the right hand side of (14) does not change when z is replaced by 1 − z, i.e.
z 1−z
−z/2
−(1−z)/2
π
Γ
ζ(z) = π
Γ
ζ(1 − z).
(33)
2
2
Now, if we define
z
1
ζ(z),
(34)
ξ(z) = z(z − 1)π −z/2 Γ
2
2
then the simple poles Γ(0) and ζ(1) will cancel with the terms z and (1 − z), so we obtain
Theorem 7 (Functional Equation) ξ(z) is an entire function, and
ξ(z) = ξ(1 − z).
10
(35)
3.2
The zeros of ζ(z)
From the Euler’s product formula (7) it follows that ζ(z) 6= 0 if ℜz > 1. If ℜz < 0 then
ℜ(1 − z) > 1 and the right hand side of (33) does not vanish. On the other hand Γ(z/2)
has simple poles in −2N := {−2, −4, −6, · · ·}. Therefore ζ(z) must have first order zeros
in −2N. These are the so-called trivial zeros of ζ(z). Now the non-trivial zeros must lie in
the closure of the critical strip 0 < ℜz < 1. The question of the location of this zeros is
far more complicated, and in fact is still unsolved. We begin with a prove that this zeros
should be complex.
Lemma 2 Let ℜz > 0, then ζ(z) is given by
(1 − 21−z )ζ(z) = 1 −
1
1
1
+ z − z ···.
z
2
3
4
(36)
Proof: If ℜz > 1, then
1−z
(1 − 2
∞
∞
X
X
1
1
1
1
1
)ζ(z) =
−2
= 1 − z + z − z ···.
z
z
n
(2n)
2
3
4
n=1
n=1
The last series converges for z in ℜz > 0 by Abel-Dirichlet-Dedekind generalization of the
alternating series test (see [18], §5.5):
P∞
If bn → 0 and {bn } is sequence of bounded variation, i.e.
n=1 |bn − bn+1 | < ∞, then
∞
X
(−1)n bn − converges.
n=1
In our case {n−z } is of bounded variation since
1
1
1
=O
−
,
nz
(n + 1)z n1+ℜz
(37)
so the equality (36) follows from the uniqueness of the analytical continuation.
Corollary 2 If z > 0 then ζ(z) > 0.
Lets multiply (33) by z and take the limit z → 0. We have
z Γ(z/2) → 2,
so it follows that 2ζ(0) = −1, i.e.
zζ(1 − z) = z −z −1 + · · · → −1,
1
ζ(0) = − .
(38)
2
From (36) and (38) we can deduce that if 0 ≤ z < 1, then ζ(z) 6= 0. Therefore the zeros
located in the strip 0 ≤ ℜz ≤ 1 should be complex numbers. Moreover, from the definition
(34) it follows that the poles of Γ(z/2) will cancel with the zeros of ζ(z) in −2N, so the
zeros of ξ(z) will be exactly the complex zeros of ζ(z). Finally, from (34) and (32) it is clear
that
ξ(z) = ξ(z̄)
(39)
This plus (35) implies that if ρn is a zero of ξ(z) so are ρn , 1 − ρn and 1 − ρn . In summary,
we have proved:
11
Theorem 8 The zeros of ζ(z) are the even negative integers −2N and complex numbers
{ρn }, which all lie in the closure of the critical strip 0 < ℜz < 1 and are situated symmetrically with respect to the lines ℑz = 0 and ℜz = 1/2. The complex zeros of ζ(z) are exactly
the zeros of ξ(z).
Our next result will show that there are no zeros on the boundary of the critical strip. As we
already mentioned, Wiener has proved, that this statement is equivalent to the prime number
theorem (see [14]). The idea of the proof depends on the following simple trigonometric
inequality
3 + 4 cos(α) + cos(2α) ≥ 0,
and a series expansion of the logarithmic derivative of ζ(z). The coefficients in this series
will be given by a number-theory function, known as the von Mangoldt symbol:
log p if n = pk for some p ∈ P, k ∈ N
Λ(n) =
(40)
0
otherwise.
Lemma 3 For ℜz > 1,
∞
ζ ′ (z) X Λ(n)
=
.
−
ζ(z)
nz
(41)
n=1
Proof: We will use the following fact (proved as a Lemma in homework #5):
If fn : G 7→ C are
Q holomorphic and non-vanishing functions in the domain G, and the
product f (z) = ∞
n=1 fn (z) converges uniformly on compact subsets of G, then
∞
f ′ (z) X fn′ (z)
=
f (z)
f (z)
n=1 n
uniformly on compact subsets of G.
We apply this for the Euler’s product formula (7) and then rearrange the series (which is
correct, because of the absolute convergence):
−
∞
∞
k=1
n=1
X X log p X Λ(n)
ζ ′ (z) X (1 − p−z )′ X p−z log p X log p
=
=
=
=
=
.
−z
−z
z
ζ(z)
(1 − p )
1−p
p −1
p kz
nz
p
p
p
p
(42)
Lemma 4 Suppose Φ(z) 6≡ 0 is a holomorphic
in a neighborhood of P ∈ R. If
function
Φ′ (z)
Φ(P ) = 0, then there exists ǫ > 0 such that ℜ Φ(z) > 0 for P < z < P + ǫ.
Proof: Let Φ has a zero of order k ≥ 1 in P . In the neighborhood of P we have the
expansion:
Φ(z) = α(z − P )k + · · · =⇒ Φ′ (z) = kα(z − P )k−1 + · · · .
(43)
Therefore
ℜ
Φ′ (z)
Φ(z)
= ℜ k(z − P )−1 + · · · > 0
in some neighborhood of P of the given form.
12
(44)
Theorem 9 ζ(z) has no zeros on the boundary of the critical strip.
Proof: Suppose that ∃ t0 ∈ R−{0}, such that ζ(1 + it0 ) = 0. Define
Φ(z) = ζ 3 (z) ζ 4 (z + it0 ) ζ(z + 2it0 ).
(45)
At z = 1 the third order pole of Φ(z) cancels with (at least) the fourth order zero, so Φ(z)
is holomorphic in a neighborhood of 1, and Φ(1) = 0. By Lemma 4 we have
′ Φ (z)
ℜ
> 0 , for 1 < z < 1 + ǫ.
Φ(z)
On the other hand, for 1 < z < 1 + ǫ we have Φ(z) 6= 0, and direct calculations show
Φ′ (z)
Φ(z)
ζ ′ (z)
ζ ′ (z + it0 ) ζ ′ (z + 2it0 )
+4
+
ζ(z)
ζ(z + it0 )
ζ(z + 2it0 )
∞
X
Λ(n) n−z −3 − 4n−it0 − n−2it0 ,
=
= 3
n=1
that is
ℜ
Φ′ (z)
Φ(z)
= −
= −
∞
X
n=1
∞
X
n=1
n
o
Λ(n) n−z 3 + 4 cos(t0 log n) + cos(2t0 log n)
o
n
Λ(n) n−z 2 (cos(t0 log n) + 1)2 ≤ 0.
This is a contradiction, so ζ(1 + it0 ) 6= 0, ∀t0 ∈ R−{0}. From the functional equation (35)
it follows that ζ(it0 ) 6= 0, ∀t0 ∈ R−{0}, also ζ(0) 6= 0, by (38).
A lot of work has been done studying the zeros of ζ(z) in the critical strip. One direction of
research involves N (T ) - the number of zeros of ζ(z) in the rectangle {0 < ℜz < 1} ∩ {0 <
ℑz ≤ T }. Since ζ(z) 6≡ 0, N (T ) < ∞. Riemann conjectured in [29], and von Mangoldt proved
in 1895, that
T
T
T
log
+ O(log T ).
(46)
−
N (T ) =
2π
2π
2π
In 1914, Godfrey Harold Hardy (1877-1947) obtained that an infinite number of the zeros of
ζ(z) do occur on the critical line. In 1974 Levin showed in [22] that at least 1/3 (as a ratio
to N (T )) of the roots must lie on ℜz = 1/2, a result which has since been sharpened to
40% by Conrey in 1989 (see [34]). Computational results from 1986 show, that the first
1, 500, 000, 001 non-trivial zeros do indeed have real part one-half (see [24]). However, the
main conjecture in the field, that there are no other roots, stays unproven.
Another interesting point is the study of ζ(z) at the integers. Euler computed ζ(2k) for
k = 1 . . . 13:
π2
π4
ζ(2) =
, ζ(4) =
, . . . (see homework #4).
6
90
For n - even integer it is known, that ζ(n) is transcendental and can be represented as
ζ(n) =
2n−1 π n |Bn |
,
n!
13
where Bn are the Bernoulli numbers defined by the identity
∞
X Bn x n
x
=
.
ex − 1
n!
n=0
The study of the function at the odd integers is significantly more difficult. Apéry produced
the only known result in 1979 which states that ζ(3) is irrational (see [1]).
4
The Prime Number Theorem
“The shortest path between two truths in the real domain passes through the
complex domain.”
J. Hadamard
In this section we give a proof of (1), following the approach from [26] (see also [25]). We
begin with some definitions from the number theory and prove a theorem of Chebyshev in
§4.1. In §4.2, we give a convergence result that is used later in §4.3, where an outline of the
proof is presented.
4.1
Chebyshev’s Theorem
The statement of the prime number theorem suggests that the asymptotic behavior of the
primes P
should be considered with weight log(·). This gives us a motivation to replace
π(x) = p≤x 1 with the following functions, defined by Chebyshev:
θ(x) =
X
log p
and ψ(x) =
X
log p =
pm ≤x
p≤x
X
Λ(n),
n≤x
where Λ(x) is the von Mangoldt symbol (40).
Lemma 5 For every x ≥ 3 the following inequalities are true:
log x
1
ψ(x)
log x
ψ(x)
≤ π(x)
≤
+
.
x
x
log x
x
log x − 2 log log x
Proof: For p ∈ P let kp = max{k ∈ N : pk ≤ x}. Since kp log p ≤ log x,
ψ(x) =
X
p≤x
kp log p ≤ π(x) log x =⇒
ψ(x)
log x
≤ π(x)
.
x
x
If 0 < y < x, then
π(x) = π(y) +
X
y<p≤x
1≤y+
X
1
1
log y ≤ y +
ψ(x).
log y
log y
y<p≤x
Take y = x/ log2 x; y < x for x ≥ 3. The lemma follows from the inequality
π(x) ≤
x
ψ(x)
.
+
2
log x log x − 2 log log x
14
(47)
Corollary 3 The statement of the prime number theorem is equivalent to
ψ(x)
= 1.
x→∞ x
lim
Next we show that if the above limit exists its value should be 1.
Lemma 6
log n =
X
Λ(j)
j|n
Proof: Let n = pk11 · · · pkmm . The only non-zero terms on the right are log p1 , · · · , log pm .
Moreover log P
pi appears only for j = pi , · · · , pki i i.e. exactly ki times. The identity follows
from log n = m
i=1 ki log pi .
Lemma 7
n
X
n
ψ
log n! =
j
j=1
Proof: By the previous Lemma and the definition (47):
log n! =
n
X
log k =
n X
X
X
Λ(i) =
Theorem 10
lim inf
x→∞
n X
X
Λ(i) =
n
X
n
ψ
.
j
j=1
j=1 i≤n/j
ij≤n
k=1 ij=k
k=1
Λ(i) =
ψ(x)
ψ(x)
≤ 1 ≤ lim sup
x
x
x→∞
(48)
Proof: The idea of this proof is borrowed from [27]. Fix m and define
ψ(x)
,
x≥m x
L−
m = inf
ψ(x)
,
x≥m x
L+
m = sup
ψ(x)
< ∞.
1≤x<m x
Mm = sup
(49)
+
It will be sufficient to show that L−
m ≥ 1, and Lm ≤ 1. Our starting point is the expression
from the previous Lemma:
n
n
log n! X ψ(n/j) X 1 ψ(n/j)
=
=
.
n
n
j n/j
j=1
(50)
j=1
Consequently
[n/m]
L−
m
[n/m]
X 1
X 1
log n!
≤
≤ L+
+ Mm
m
j
n
j
j=1
j=1
n
X
j=[n/m]+1
1
.
j
At this point we need some well-known asymptotic results. First of all
( n
)
n
X1
X
1
= log n + O(1) , since lim
− log n = γ − Euler’s constant.
n→∞
j
j
j=1
j=1
15
(51)
(52)
Also, from the Stirling’s formula
√
√
nn e−n 2πn < n! < nn e−n+1/(12n) 2πn
(53)
it follows that
log n! = n log n + O(n).
(54)
Plugging (54) and (52) back in (51) yields
L+
m (log n
L−
m (log n − log m + O(1)) ≤ log n + O(1),
− log m + O(1)) + Mm (log m + O(1)) ≥ log n + O(1).
(55)
(56)
This holds for every n > m, so
log n + O(1)
log n − log m + O(1)
log n + O(1) − Mm (log m + O(1))
≥
log n − log m + O(1)
L−
m ≤
L+
m
→ 1,
n→∞
(57)
→ 1.
n→∞
(58)
+
Hence L−
m ≥ 1, and Lm ≤ 1, which completes the proof.
Corollary 4
π(x) ∼
ψ(x)
x
⇐⇒ ∃ lim
x→∞ x
log x
.
4.2
Convergence Theorem
Theorem 11 Consider a sequence {an }∞
n=1 ⊂ C which satisfy |an | ≤ 1. Clearly the series
P
∞
−z
converges absolutely and uniformly on the half-plane ℜz > 1. The limit funcn=1 an n
tion F (z) should be holomorphic only for ℜz > 1, but suppose that itPcan be continued to a
−z converges
holomorphic function on a neighborhood of ℜz ≥ 1. Then the series ∞
n=1 an n
also for ℜz ≥ 1.
Proof: Fix w such that ℜw ≥ 1. The function G(z) = F (z + w) is holomorphic for ℜz ≥ 0.
This means that if we take any R, R ≥ 1 then there exist δ, δ ≤ 1/2 such that G(z) is
holomorphic in ΩR,δ , where ΩR,δ = {ℜz > −δ} ∩ {|z| < R} (see fig.2). Denote with Γ the
−δ
(0,0)
Γ= B
R
+
A
Figure 2: The domain ΩR,δ , and its boundary Γ.
16
boundary of ΩR,δ equipped with counterclockwise orientation. Let M = supz∈ΩR,δ |G(z)|,
then
|F (z + w)| ≤ M , z ∈ Γ.
(59)
Consider the integral
I1 =
Z
F (z + w)N
Γ
z
z
1
+
dz,
z R2
(60)
where N will be a “large” integer. The function F (z + w)N z z −1 is meromorphic in ΩR,δ ,
with simple pole at z = 0. By the Residue Theorem
I1 = 2πi Res F (z + w)N z z −1 , 0 = 2πi F (w).
(61)
Split
A + B, where A = Γ ∩ {ℜz > 0} and B = Γ P
∩ {ℜz < 0}. Denote SN (z) =
PN Γ =−z
−z - the remainder
- the partial sum of the series, and rN (z) = ∞
k=1 an n
n=N +1 an n
term. Observe that on A one has F (z+w) = SN (z+w)+rN (z+w), since A ⊂ {ℜz > 0}. On
the other hand SN (z) is polynomial, and therefore entire function. Thus SN (z+w)N z (z −1 +
zR−2 ) will be meromorphic in C with one simple pole in z = 0. We can apply again the
residue theorem, this time for the contour C = C(0, R) = {|z| = R}:
Z
z
z 1
SN (z + w)N
+
I2 =
dz = 2πi SN (w).
(62)
z R2
C
Observe that C = A + −A, where −A = {−z : z ∈ A}, so
Z
Z
1
1
z
z
SN (z + w)N z
I2 =
SN (z + w)N z
+ 2 dz +
+ 2 dz
z R
z R
A
−A
Z z
1
+ 2 dz.
=
SN (z + w)N z + SN (w − z)N −z
z R
A
Subtracting this from (60) gives:
Z
Z 1
z
z
1
F (z + w)N z
+ 2 dz +
+ 2 dz.
I1 − I2 =
rN (z + w)N z − SN (w − z)N −z
z R
z R
B
A
Let IA and IB be the integrals on the right. Our goal is to prove that |F (w) − SN (w)| → 0,
when N → ∞. One way to obtain this difference is to subtract (62) from (61), so
2π |F (w) − SN (w)| ≤ |I1 − I2 | ≤ |IA | + |IB |.
(63)
Now, if we can estimate IA and IB (such that they vanish to 0 as N → ∞) then the theorem
will follow. Denote x = ℜz, y = ℑz. We will estimate separately each term in IA and IB :
z
z̄
z
2x
1
+ 2 = 2 + 2 = 2 , on {|z| = R},
z R
|z|
R
R
2
2
1
2
+ z ≤ 1 + 1 |z| ≤ 1 1 + |z|
, on {x = −δ, |z| ≤ R},
≤
z R2 |z| |z| R2
2
δ
R
δ
|rN (z + w)| ≤
∞
X
n=N +1
∞
∞
X
X
1
1
|an |
≤
≤
.
z+w
x+ℜw
x+1
|n
|
n
n
n=N +1
17
n=N +1
(64)
(65)
(66)
Since for x > −1 the function n−x−1 decreases it follows that the area below the graph of
n−x−1 is bigger then the Riemann sum on the right-hand side of (66), (see also fig.1). Thus
Z ∞
1
n−x ∞
dn
= x
=−
, for x > −1.
(67)
|rN (z + w)| ≤
x+1
x N
N x
N n
|SN (w − z)| ≤
N
X
n=1
|an | |n
z−w
|≤
N
X
n
x−ℜw
≤
n=1
N
X
nx−1 .
(68)
n=1
The following estimates are obtained by a similar geometric idea as in fig.1:
N
X
n=1
nx−1 ≤
Z
1
N +1
nx−1 dn for x ≥ 1, and
N
X
n=1
nx−1 ≤
Z
N
0
nx−1 dn for x ≤ 1.
So, for any x ∈ R:
|SN (w − z)| ≤ N
x−1
+
Z
N
n
x−1
dn = N
0
x
1
1
+
.
N
x
(69)
Now using (64), (67), and (69) we can apply “maximum times length” estimate for IA :
(
)
1 1 2x
1
x
x 1
N +N
+
πR,
(70)
|IA | ≤ max
0≤x≤R
N xx
N
x N x R2
i.e.
|IA | ≤ max
0≤x≤R
(
1
2
+
x N
2x
R2
)
πR ≤
4π 2π
+
.
R
N
(71)
In order to estimate IB , we introduce the curves: B1 R= {ℜz = −δ, |z| ≤ R} and B2 = {0 ≥
ℜz ≥ −δ, |z| = R}. Clearly B = B1 + B2 . Let IBj = Bj F (z + w)N z (z −1 + zR−2 ) dz, then
|IB | ≤ |IB1 | + |IB2 |.
Using (59) and (65) we can estimate IB1 :
√
R2 −δ2
Z R
2
2
4M R
|IB1 | ≤ √
M N −δ dy =
MN
dy ≤
.
δ
δ
δN δ
2
2
−R
− R −δ
o
n
√
A parameterization of B2 is given by z = x ± i R2 − x2 , x ∈ (−δ, 0) . Hence
Z
−δ
(72)
s
r
r
r
2
2
dz 3
2
R
1
1
x
= 1+
=
=
≤
≤√ ≤ .
dx 2
2
2
2
2
2
2
R −x
R −x
1 − x /R
1−δ
2
3
Using this plus (59) and (64)
Z 0
2|x|
MNx 2
|IB2 | ≤ 2
R
−δ
i.e.
|IB2 | ≤
produces the following estimate for IB2 :
Z
3
6M 0
6M N δ − δ ln(N ) − 1
dx = − 2
xN x dx = 2
,
2
R −δ
R
log2 (N )N δ
6M
R2 log2 (N )
=⇒ |IB | ≤
18
4M R
6M
.
+ 2 2
δ
δN
R log (N )
(73)
(74)
Now we can put all together. The substitution of (74) and (71) in (63) gives (π > 3):
|F (w) − SN (w)| ≤
1
MR
M
2
+
+
+ 2 2
.
δ
R N
δN
R log (N )
(75)
When w is fixed in ℜw ≥ 1 this estimate is true for any R ≥ 1 and N ∈ N. Let 3 > ǫ > 0
be arbitrary “small” real number. Take R = R(ǫ) = 3/ǫ > 1. Take N = N (ǫ) sufficiently
big such that N > N (ǫ) implies
ǫ
MR
M
1
< ,
+
+ 2
2
δ
N
δN
3
R log (N )
this is possible, since δ = δ(R) ≤ 1/2 and M = M (R, δ) are fixed when R is fixed. Now,
plugging this estimate back in (75) gives that if N > N (ǫ) then
ǫ
ǫ
|F (w) − SN (w)| ≤ 2 + = ǫ.
3 3
This is the very definition for convergence.
4.3
A Proof of the Prime Number Theorem
Let d(n) denotes the number of divisors of n, and µ(n) be the Möbius function (8). For
ℜz > 1 we have the following properties of ζ(z):
∞
X
d(n)
ζ 2 (z) =
n=1
−ζ ′ (z) =
nz
,
∞
X
log n
n=1
nz
(76)
,
(77)
∞
X µ(n)
1
=
,
ζ(z) n=1 nz
(78)
Indeed, to prove (76) observe that if ℜz > 1, then
X
∞
∞
∞ ∞
X
X
1 X 1
d(n)
1 X
ζ (z) =
1 =
=
.
z
z
z
i
j
n
nz
2
i=1
j=1
n=1
ij=n
n=1
The series multiplication is justified by the absolute convergence.
Next, (77) follows from
∞
∞
X
X
log n
.
(n−z )′ = −
ζ ′ (z) =
nz
n=1
n=1
And (78) from
X
∞
Y
1
µ(n)
1
=
1− z =
.
ζ(z)
p
nz
p
n=1
Again this operation is correct, since the series converges absolutely.
19
Given (76)-(78) we can outline our proof of the prime number theorem:
• By Theorem 9, ζ(z) 6= 0 in ℜz ≥ 1, so the function 1/ζ(z) is holomorphic in
ℜz
|µ(n)| ≤ 1 we can apply the convergence Theorem 11 for the series
P∞≥ 1. Since
−z , and conclude that it converges in ℜz ≥ 1.
µ(n)n
n=1
• In particular (78) will hold also in the limit case z = 1, i.e.
∞
X
µ(n)
n
n=1
= 0.
(79)
This will be our main identity. It can be proved (see [2]), that it is equivalent to the
prime number theorem.
• By Corollary 3, the prime
P number theorem is equivalent to the statement ψ(x) ∼ x,
which is the same as n≤N Λ(n) ∼ N . In other
words (see (41)), the coefficients an
P
z
in the series representation −ζ ′ (z)/ζ(z) = ∞
n=1 an /n should tend to 1 in average.
This means that the coefficients bn corresponding to
−
∞
X bn
ζ ′ (z)
− ζ(z) =
ζ(z)
nz
(80)
n=1
should vanish in average, since bn = an − 1.
• Combining (76), (77) and (78) we get
∞
∞
∞
X
ζ ′ (z)
1 ′
µ(n) X log n X d(n)
2
−
− ζ(z) =
−
−ζ (z) − ζ (z) =
,
ζ(z)
ζ(z)
nz
nz
nz
n=1
n=1
therefore
X
bn =
n≤N
X
ij≤N
n=1
h
i
µ(i) log j − d(j) .
(82)
• The idea is that we can find the asymptotic of each term in (82):
PN µ(i)
◦ By (79):
i=1 i = o(1 )
√
PN µ(i)
√ = o( N )
◦ Corollary 6:
i=1
i
P
√
◦ Corollary 5: ∃γ1 ∈ R such that nj=1 log j − d(j) = γ1 n + O( n).
We will prove this corollaries after the outline.
P
• Using this results we can show that n≤N bn = o(N ):
X
bn =
i=1
n≤N
so
N
X
µ(i)
[N/i] h
X
j=1
"
r !#
N
i X
N
N
,
µ(i) γ1 + O
log j − d(j) =
i
i
N
X
µ(i)
1 X
bn = γ1
+O
N
i
n≤N
i=1
i=1
N
1 X µ(i)
√
√
N i=1 i
20
!
→0
(81)
, as N → ∞.
Lemma 8
n
X
m=1
√
d(m) = n log n + (2γ − 1)n + O( n).
(83)
Proof: Clearly
n
X
d(m) =
n X
X
1=
m=1 d|m
m=1
n [n/d]
X
X
1=
d=1 k=1
n X
n
d=1
d
.
Geometrically (see fig.3) this is the number of lattice points (x, y) ∈ N2 on, or bellow the
hyperbola xy = n. Indeed, for fixed x ∈ [1, n] the number of y ∈ N such that 1 ≤ xy ≤ n
is exactly [n/x]. By symmetry this number is equal to twice the number of the points with
y
xy = n
( n, n )
x
(0,0)
Figure 3: The lattice points (x, y) ∈ N2 below xy = n
y > x plus the number of the points with x = y, i.e.
√ √
n
]
[
n
]
+
1
[
X
√
√
1
n
d(m) = 2
−2
+ O( n) =
− x + [ n ] = 2n
x
x
2
m=1
x=1
x=1
√
√
√
1
− n + O( n) = n log n + (2γ − 1) n + O( n).
= 2n log n + γ + O √
n
n
X
√
[ n ] X
√
[ n]
Corollary 5 Using the Stirling’s formula (53) we can choose γ1 ∈ R, such that
Pn √
log
j
−
d(j)
= γ1 n + O( n).
j=1
Lemma 9
∞
X
an
n
n=1
Proof: Let PN :=
PN
an
n=1 n
= 0 =⇒ SN :=
N
X
√
a
√n = o( N ).
n
n=1
be the partial sum of the series on the left-hand side. We have
SN − a1 =
=
N
N
N
X
X
√ an X
√
a
√n =
n
n (Pn − Pn−1 )
=
n
n
n=2
n=2
n=2
N
−1
X
n=2
√
√
√
√
( n − n + 1) Pn − 2 P1 + N PN .
21
Therefore
N −1
√
SN
1 X √
lim √ = lim √
( n − n + 1) Pn .
N →∞
N →∞
N
N n=2
(84)
Recall the Stolz’s theorem:
If {an }, {bn } are sequences of real numbers and {bn } is monotonically increasing, then
an
an+1 − an
= L =⇒ ∃ lim
= L,
n→∞ bn
n→∞ bn+1 − bn
√
P
√
and note the following implication (bn = N , an = N
n=1 cn / n):
lim
lim cn = 0 =⇒
n→∞
N
1 X cn
√ = 0.
lim √
N →∞
N n=1 n
The Lemma follows from the observation that in (84) we can set
√
√
√ √
− n
√
Pn → 0 , as n → ∞.
cn = n ( n − n + 1) Pn = √
n+ n+1
Corollary 6
(79) =⇒
N
X
√
µ(n)
√ = o( N ).
n
n=1
22
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