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A simple proof of the Prime Number Theorem is constructed from Mean Value Theorem for arithmetic functions, and basic properties of the zeta function. This proof does not require any deep knowledge of the prime numbers, and it does not require any difficult to prove zerofree region beyond the well known, and simplest zerofree region {s ∈ C : Re(s) > 1}. In addition, a similar proof of the Dirichlet Theorem for primes in arithmetic progressions is included. Theorem 1.1. (Prime Number Theorem) For all sufficiently large number x ≥ 1, the prime counting function satisfies the asymptotic formula x x π(x)=#{p ≤ x : p prime} = + o . (1) log x log x   Proof. For n ≥ 1, let f(n) = Λ(n) be the weighted characteristic function of the prime numbers, see the definition (13). The corresponding is ′ ′ Λ(n) ζ (s) ζ (s) = − = −ζ(s) · (2) ns ζ(s) ζ2(s) nX≥1 for s ∈ C with Re(s) > 1, see Lemma 2.1. By Lemma 2.2, the function ′ ζ (s) g(n) = (3) ζ2(s) ns nX≥1 converges at s = 1. Therefore, by Theorem 3.1 or Theorem 3.3, the mean value of f(n) = Λ(n), in Section 3, is given by ′ 1 ζ (1) M(f) = lim Λ(n)= − = 1, (4) x→∞ x ζ2(1) nX≤x see equations (20) and (21). These immediately imply that

Λ(n)= x + o(x). (5) nX≤x Next, let ψ(t)= n≤t Λ(n). Then, by partial summation, it follows that P Λ(n) π(x)= 1 = + O x1/2 log2 x log n Xp≤x nX≥x   x 1 = dψ(t)+ O x1/2 log2 x (6) log t Z2 x x   = + o , log x log x   A Simple Proof Of The Prime Number Theorem 3 here the error term O x1/2 log2 x accounts for the prime powers pk with k ≥ 2.   This is probably one of the simplest proof of the Prime Number Theorem, there is a claim for the simplest proof in [2, p. 80].

1.1 Extension to Dirichlet Theorem As the mean value theorem for arithmetic functions has a natural extension to arithmetic progression, it is natural to consider the next result. Theorem 1.2. (Dirichlet Theorem) For all sufficiently large number x ≥ 1, and a pair of integers a ≥ 1 and q ≥ 1, with gcd(a, q) = 1, the prime counting function satisfies the asymptotic formula 1 x x π(x, q, a)=#{p ≤ x : p ≡ a mod q}≥ (1 + o (1)) + o . q log x q log x  (7) Proof. Assume q ≥ 2 is prime, and consider the Mobius pair

f(n)= g(d) and g(n)= −µ(n) log n. (8) Xd|n By Theorem 4.1, the mean value of the arithmetic function f(n) = Λ(n) over arithmetic progression {qn + a : gcd(a, q)=1 and n ≥ 0} is given by 1 M(f) = lim Λ(n) x→∞ x n≤x, nX≡a mod q 1 c (a) g(dn) = d q d n Xd|q nX≥1 1 c (a) µ(dn) log(dn) = − d q d n Xd|q Xn≥1 c (a) µ(n) log n 1 c (a) µ(dn) log(dn) = − 1 − d q n q d n Xn≥1 1X

The last line follows from Lemma 2.2, and c1(a) = 1. This and Theorem 3.3 immediately imply that 1 x x π(x, q, a) ≥ (1 + o (1)) + o , (10) q log x q log x   where M(f) = 1/q + o(1/q). This proves the claim. 

Remark 1.1. The proof of Theorem 4.1 has no limitations on the range of values of q ≥ 1. Thus, this result is probably an improvement on the Siegel-Walfisz Theorem, which states that

1 x β π(x, q, a)= + O xe−c(log log x) , (11) ϕ(q) log x   where q = O(logB x), with B > 0, c > 0 is an absolute constant, and 0 <β< 1 constants.

2 Powers Series Expansions of the Zeta Function

Let N = {0, 1, 2, 3,...} be a nonnegative integer. The Mobius function is defined by 1 if n is squarefree, µ(n)= (12) (0 if n is not squarefree. And the vonMangoldt function is defined by

log p if n = pk, k ≥ 1, is a , Λ(n)= (13) (0 if n 6= pk, k ≥ 1, is not a prime power.

−s Lemma 2.1. Let Λ be the vonMangoldt function, and let ζ(s)= n≥1 n be the zeta function. Then, P ′ ζ (s) Λ(n) − = (14) ζ(s) ns nX≥1 is absolutely convergent on the complex half plane Re(s) > 1. Proof. Since the zeta function is convergent on the complex half plane {s ∈ C : Re(s) > 1}, and it has an Euler product

1 1 −1 ζ(s)= = 1 − . (15) ns ps nX≥1 pY≥2   A Simple Proof Of The Prime Number Theorem 5

Taking the logarithm derivative yields d d 1 log ζ(s) = − log 1 − (16) ds ds ps Xp≥2   d 1 = − ds npns Xp≥2 nX≥1 log p = pns Xp≥2 nX≥1 Λ(n) = − . ns nX≥1 The second line follows from the absolute converges on the complex half plane Re(s) > 1, rearranging the double sums, and the definition of the vonMangoldt function in equation (13). 

−s The zeta function ζ(s)= n≥1 n is an entire function of a complex vari- able s ∈ C with a simple pole at s = 1. At each fixed P s0 ∈ C, it has a Taylor series expansion of the form 1 ζ(n)(s ) ζ(s)= − + 0 (s − s )n, (17) s − 1 n! 0 nX≥0 where ζ(n)(s) is the nth derivative.

The expansions at the zeros, and some other special values of the zeta function ζ(s) are simpler to determine, for example, s0 ∈ Z = {−2n : n ≥ 0}. The well known expansion at s = 1 has the power series 1 γ ζ(s) = − + n (s − 1)n (18) s − 1 n! Xn≥0 1 1 = − + γ − γ (s − 1) + γ (s − 1)2 + · · · , s − 1 0 1 2 1 see where γn is the nth Stieltjes constant, see [8], and [3, Eq. 25.2.4]. And the expansion at s = 0 has the power series 1 ζ(s)= − + (−1)nδ sn, (19) s − 1 n nX≥0 see [3]. The power series expansions at the zeros s0 are some sort of ‘zeta modular forms’ since the first coefficient a0 = ζ(s0) = 0. A Simple Proof Of The Prime Number Theorem 6

′ Lemma 2.2. The complex value function −ζ (s)/ζ2(s) is analytic and ze- rofree on the complex half plane ℜe(s) ≥ 1, and at s = 1.

Proof. Since the zeta function is zerofree on the complex half plane ℜe(s) > 1, it is sufficient to consider it at the simple pole at s = 1. Toward this end, use the power series (18) to rewrite the function as a ratio

′ −2 ζ (s) −(s − 1) − γ1 + γ2(s − 1) + · · · − 2 = −2 2 . (20) ζ (s) (s − 1) − γ1(s − 1) + γ2(s − 1) + · · · Taking the limit yields

′ ′ ζ (1) (s − 1)2ζ (s) = lim = −1. (21) ζ2(1) s→1 (s − 1)2ζ2(s)

This shows that is well defined at each point on the complex half plane ℜe(s) > 1, and at s = 1. 

Lemma 2.3. Let µ, Λ : N → C be the Mobius, and the vonMangoldt arith- metic functions. Then

µ(d)Λ(n/d)= o(x). (22) nX≤x Xd|n Proof. By Lemma 2.1 and Lemma 2.2, the series

′ ζ (s) 1 Λ(n) µ(n) − · = − · (23) ζ(s) ζ(s) ns ns nX≥1 nX≥1 d|n µ(d)Λ(n/d) = − s P n nX≥1 f(n) = ns nX≥1 converges on the complex half plane ℜe(s) > 1, and at s = 1. Therefore, by Lemma 3.1, the summatory function

f(n)= µ(d)Λ(n/d)= o(x). (24) nX≤x nX≤x Xd|n for large x ≥ 1.  A Simple Proof Of The Prime Number Theorem 7

3 Mean Values of Arithmetic Functions

Let f : N → C be a complex valued arithmetic function on the set of nonnegative integers. The mean value of an arithmetic function is defined by 1 M(f) = lim f(n). (25) x→∞ x nX≤x The mean value of an arithmetic function is sort of a weighted density of the subset of integers supp(f)= {n ∈ N : f(n) 6= 0} ⊂ N, which is the support of the function f, see [17, p. 46] for a discussion of the mean value. The natural density of a subset of integers A⊂ N is defined by

{n ≤ x : n ∈ A} δ(A) = lim . (26) x→∞ x

3.1 Some Results on Mean Values of Arithmetic Functions This section investigates the two cases of , and divergent series f(n) f(n) < ∞ and = ±∞, (27) n n Xn≥1 nX≥1 and the corresponding mean values 1 1 M(f) = lim f(n)=0 and M(f) = lim f(n) 6= 0 x→∞ x x→∞ x nX≤x nX≤x respectively. First, a result on the case of convergent series is considered here.

Lemma 3.1. Let f : N −→ C be an arithmetic function. If the series −1 n≥1 f(n)n converges, then its mean value P 1 M(f) = lim f(n) = 0 (28) x→∞ x nX≤x vanishes.

−1 Proof. Consider the pair of finite sums n≤x f(n), and n≤x f(n)n . By hypothesis, f(n)n−1 = c + o(1), where c 6= 0 is constant, for large n≤x P P P A Simple Proof Of The Prime Number Theorem 8 x ≥ 1. Therefore f(n) f(n) = n · (29) n n≤x n≥1 X Xx = t · dR(t) 1 Z x = xR(x) − R(1) − R(t)dt Z1 = o(x),

−1 where R(t) = n≤t f(n)n . By the definition of the mean value of a function in (25), this confirms that f(n) has mean value zero.  P This is standard material in the literature, see [13, p. 4]. The second case is covered by a few results on divergent series, which are considered next.

−1 Let f(n) = d|n g(d), and let the series c = n≤x g(n)n be absolutely convergent. Under these conditions, the mean value of the function f(n) P P can be determined indirectly from the properties of the function g(n).

Theorem 3.1. (Wintner) Consider the arithmetic functions f, g : N −→ C, and assume that the associated generating series are zeta multiple

f(n) g(n) = ζ(s) . (30) ns ns nX≥1 Xn≥1 Then, the followings hold.

−s (i) If the series n≥1 f(n)n is defined for ℜe(s) > 1, and the series c = g(n)n−1 is absolutely convergent, then the mean value, and n≤x P the partial sum are given by P 1 g(n) M(f) = lim and f(n)= cx + o(x). x→∞ x n nX≥1 nX≤x

−s (ii) If the series n≥1 f(n)n is defined for ℜe(s) > 1/2, and the series c = g(n)n−1 is absolutely convergent, then, for any ε > 0, the n≤x P mean value, and the partial sum are given by P 1 g(n) M(f) = lim and f(n)= cx + O x1/2+ε . x→∞ x n nX≥1 nX≤x   A Simple Proof Of The Prime Number Theorem 9

Proof. (i) The partial sum of the series (30) is rearranged as

f(n) = g(d) (31) nX≤x nX≤x Xd|n = g(d) 1 Xd≤x nX≤x/d x x = g(d) − d d Xd≤x  n o g(d) = x + O |g(d)| , d   Xd≤x Xd≤x   wheere {x} = x−[x] is the fractional part function. The first line arises from −s −s the of the power series ζ(s) = n≥1 n , and n≥1 g(n)n . This is followed by reversing the order of summation. Moreover, the first P P finite sum is g(n) g(n) = + o(1) = c + o(1). (32) n n nX≤x nX≥1 −1 because n≥1 g(n)n is absolutely convergent. A use a dyadic method to split the second finite sum as P |g(d)| |g(d)| |g(d)| = · d + · d (33) d d 1/2 1/2 Xd≤x d≤Xx x X≤d≤x |g(d)| |g(d)| ≤ x1/2 + x d d 1/2 1/2 d≤Xx x X≤d≤x = O x1/2 + o(x) = o(x).  −1  Again, this follows from the absolute convergence n≥1 |g(n)|n < ∞. Similar proofs appear in [14, p. 138], [2, p. 83],P and [6, p. 72]. Another derivation of the Wintner Theorem from the Wiener-Ikehara Theorem is also given in [14, p. 139]. Theorem 3.2. (Axer) Let f, g : N −→ C, be arithmetic functions and assume that the associated generating series are zeta multiple f(n) g(n) = ζ(s) . (34) ns ns nX≥1 Xn≥1 A Simple Proof Of The Prime Number Theorem 10

−s If the series n≥1 f(n)n is defined for ℜe(s) > 1, and n≤x |g(n)| = O(x) is convergent, then the mean value, and the partial sum are given by P P g(n) M(f)= and f(n)= cx + o(x). (35) n nX≥1 nX≤x The goal of the next result is to strengthen Wintner Theorem by removing the absolutely convergence condition.

Theorem 3.3. Let f, g : N −→ C, be arithmetic functions and assume that the associated generating series are zeta multiple

f(n) g(n) = ζ(s) . (36) ns ns nX≥1 Xn≥1 −s If the series n≥1 f(n)n is defined for ℜe(s) > 1/2, and the series c = g(n)n−1 is convergent, then, n≤x P P g(n) M(f)= and f(n)= cx + o(x). (37) n nX≥1 nX≤x Proof. (i) The partial sum of the series (36) is rearranged as

n · f(n) = n g(d) (38) nX≤x nX≤x Xd|n = g(d) n Xd≤x nX≤x/d x2 x2 = g(d) + o 2d 2d Xd≤x    x2 g(d) x2 g(d) = + o , 2 d  2 d  d≤x d≤x X X   wheere {x} = x−[x] is the fractional part function. The first line arises from −s −s the convolution of the power series ζ(s) = n≥1 n , and n≥1 g(n)n . This is followed by reversing the order of summation. Moreover, the first P P finite sum is g(n) g(n) = + o(1) = c + o(1). (39) n n nX≤x nX≥1 A Simple Proof Of The Prime Number Theorem 11

−1 because n≥1 g(n)n is absolutely convergent. A use a dyadic method to split the second finite sum as P |g(d)| |g(d)| |g(d)| = · d + · d (40) d d 1/2 1/2 Xd≤x d≤Xx x X≤d≤x |g(d)| |g(d)| ≤ x1/2 + x d d 1/2 1/2 d≤Xx x X≤d≤x = O x1/2 + o(x) = o(x). 

−1 Again, this follows from the absolute convergence n≥1 |g(n)|n < ∞. Lastly, but not least, the original partial sum is recovered by partial sum- P mation. 

4 Extension To Arithmetic Progressions

The mean value of theorem arithmetic functions over arithmetic progressions {qn + a : n ≥ 1} facilitates another simple proof of Dirichlet Theorem. −1 Theorem 4.1. Let f(n)= d|n g(d), and let the series n≥1 g(n)n 6= 0 be absolutely convergent. Then P P 1 1 c (a) g(dn) M(f) = lim f(n)= d . (41) x→∞ x q d n n≤x d|q n≥1 n≡aXmod q X X

i2πnx/k where ck(n)= gcd(x,k)=1 e . For the parameterP 1 ≤ a

1 x β π(x, q, a)= + O e−(log x) (43) ϕ(q) log x   where q = O logB x , with B > 0, and 0 <β< 1 constants.  A Simple Proof Of The Prime Number Theorem 12

5 Powers Sums Over Arithmetic Progressions

Let N = {0, 1, 2, 3,...} be the set of nonnegative integers, and let q|N + a = {qn + a : n ∈ N} be the arithmetic progression defined by a pair of integers a > 0, and q >≥ 1. The sums of powers over arithmetic progressions is one of the possible generalizations of the sums of powers , k?0, over the integers. A few estimates of the powers sums over arithmetic progressions are computed here.

Lemma 5.1. Let a ≥ 0 and q ≥ 1 be fixed integers. Let x ≥ 1 be a sufficiently large real number. Then 1 1 (i) n = x2 + o x2 . 2q q n≤x   n≡aXmod q 1 1 (ii) n ≥ x2 + O x . 2q q n≤x   n≡aXmod q Proof. The integers in a linear arithmetic progression are of the form n = qm + a, with 0 ≤ m ≤ (x − a)/q. Inserting this into the finite sum produces

n = q m + a 1 (44) n≤x m≤(x−a)/q) m≤(x−a)/q) n≡aXmod q X X q x − a x − a x − a = + 1 + a , 2 q q q       where [z] = z − {z} be the largest integer function. Set z = x − a, and expand the expression to obtain:

q x − a x − a x − a + 1 + a (45) 2 q q q       q z z z z z z = − − + 1 + a − 2 q q q q q q           1 z q z 2 z q z z z = z2 − z + + − + a − 2q q 2 q 2 2 q q q          1 1 = z2 + o z2 . 2q q   Replacing z = x − a back into the (45) yields the result.  A Simple Proof Of The Prime Number Theorem 13

The above estimates are sufficient for the intended applications. A sharper estimate of the form 1 1 1 q n = x2 + O x + (46) 2q q 4 n≤(Xx−a)/q   appears in [16, p. 83].

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