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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.

Then f /g has a pole of order 1 at z0 , and res(z0 , f /g) = f (z0 )/g (z0 ). Proof. Exercise. Let U be a non-empty, open subset of C and f a meromorphic function on U which is not identically zero. We define the logarithmic derivative of f by f /f. Suppose that U is simply connected and f is analytic and has no zeros on U . Then f /f has an anti-derivative h : U → C. One easily verifies that (eh /f ) = 0. Hence eh /f is constant on U . By adding a suitable constant to h we can achieve that eh = f .

Assume that for every compact subset K of U there is a constant CK < ∞ such that |fn (z)| CK for all z ∈ K, n 0. (k) Then f is analytic on U , and fn → f (k) pointwise on U for all k 1. Proof. The set U can be covered by disks D(z0 , δ) with z0 ∈ U , δ > 0, such that D(z0 , 2δ) ⊂ U . We fix such a disk D(z0 , δ) and prove that f is analytic on D(z0 , δ) (k) and fn → f (k) pointwise on D(z0 , δ) for k 1. This clearly suffices. Let z ∈ D(z0 , δ), k fn(k) (z) = 0. 10, we have k! 2πi γz0 ,2δ 1 = k!

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