By Jan-Hendrik Evertse

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3 An elementary result for prime numbers We finish this introduction with an elementary proof of a weaker version of the Prime Number Theorem. 4. We have 1 2 x log x π(x) 2 x for x log x 3. The proof is based on some simple lemmas. For an integer n = 0 and a prime number p, we denote by ordp (n) the largest integer k such that pk divides n. 5. Let n be an integer x. 1 and p a prime number. ) = j=1 n . pj Remark. This is a finite sum. Proof. We count the number of times that p divides n!. Each multiple of p that is n contributes a factor p.

Taking logarithms, we get π(2k + 1) log(k + 2) 2k log 2 + π(k + 1) log(k + 2), and applying the induction hypothesis to π(k + 1), using k + 2 > k + 1 > n/2, 2k log 2 2(k + 1) + log(k + 2) log(k + 1) (log 2 + 1)n + 1 n < <2 for n > 200. 1 Basics Given z0 ∈ C, r > 0 we define D(z0 , r) := {z ∈ C : |z − z0 | < r} (open disk), D(z0 , r) := {z ∈ C : |z − z0 | r} (closed disk). A subset U ⊆ C is called open if U = ∅ or if for every z0 ∈ C there is δ > 0 with D(z0 , δ) ⊂ U . A subset U ⊆ C is called closed if U c := C \ U is open.

X1 In case that x1 = M ,xr = N we are done. if x1 > M , then A(t) = 0 for M t < x1 x and thus, M1 A(t)g (t)dt = 0. If xr < N , then A(t) = A(xr ) for xr t N , hence N A(t)g (t)dt = A(N )g(N ) − A(xr )g(xr ). 1) this implies our Theorem. 2. Let f : Z>0 → C be an arithmetic function with the property that there exists a constant C > 0 such that | N C for every N 1. Then n=1 f (n)| ∞ −s Lf (s) = n=1 f (n)n converges for every s ∈ C with Re s > 0. More precisely, on {s ∈ C : Re s > 0} the function Lf is analytic, and for its k-th derivative we have ∞ (k) f (n)(− log n)k n−s .

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Analytic Number Theory [lecture notes] by Jan-Hendrik Evertse

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