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

By Jan-Hendrik Evertse

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27 and assume in addition that fn = −1 on U for every n 0. Then for the function F = ∞ n=0 (1 + fn ) we have ∞ F fn = . F 1 + f n n=0 Proof. Let Fm := m n=0 (1 + fn ). Then Fm → F uniformly on every compact subset of U . 28. 57 Chapter 3 Dirichlet series and arithmetic functions An arithmetic function is a function f : Z>0 → C. To such a function we associate its Dirichlet series ∞ f (n)n−s Lf (s) = n=1 where s is a complex variable. It is common practice (although this doesn’t make sense) to write s = σ + it, where σ = Re s and t = Im s.

Proof of π(x) 2x/ log x. Let again n = [x]. Since t/ log t is an increasing function of t, it suffices to prove that π(n) 2 · n/ log n for all integers n 3. We proceed by induction on n. It is straightforward to verify that π(n) 2 · n/ log n for 3 n 200. Let n > 200, and suppose that π(m) 2 · m/ log m for all integers m with 3 m < n. If n is even, then we can use π(n) = π(n − 1) and that t/ log t is increasing. Assume that n = 2k + 1 is odd. 6, we have 2k + 1 k p (k + 2)π(2k+1)−π(k+1) . 8, this leads to (k + 2)π(2k+1)−π(k+1) 22k .

Let ε > 0. Then there is N such that |fn (z) − fm (z)| < ε for all z ∈ K, m, n N . Choose m N . Then there is C > 0 such that |fm (z)| C for z ∈ K since fm is continuous. Hence |fn (z)| C + ε for z ∈ K, n N . 24. 26. let U ⊂ C be a non-empty open set, and {fn : U → C}∞ n=0 a sequence of analytic functions, converging to a function f pointwise on U and uniformly on every compact subset of U . Then fn (z) f (z) = n→∞ fn (z) f (z) lim for all z ∈ U with f (z) = 0, where the limit is taken over those n for which fn (z) = 0.

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