By Jan-Hendrik Evertse

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This implies (i) and (ii). 52 We finish with proving (iii). Our assumption implies that f has a Laurent series expansion ∞ an (z − z0 )n f (z) = n=−∞ converging on D (z0 , r). Then for z ∈ D0 (z0 , r) we have z ∈ D0 (z0 , r) and 0 ∞ ∞ an (z − z0 f (z) = f (z) = )n an (z − z0 )n , = n=−∞ n=−∞ which clearly implies (iii). 7 Analytic functions defined by integrals In analytic number theory, quite often one has to deal with complex functions that are defined by infinite series, infinite products, infinite integrals, or even worse, infinite integrals of infinite series.

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. Proof. Obvious. 27. Let U ⊂ C be a non-empty open set and {fn : U → C}∞ n=0 a sequence of analytic functions.

Consequently, n log 2 − log(n + 1) log n (x − 1) log 2 − log(x + 1) log x π(x) = π(n) 1 2 x for x log x 100. 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.

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

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