By Ionin Y. J., Shrikhande M. S.

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**Example text**

We f (x) say that f (x) is little oh of g(x) and write f (x) = o(g(x)) if lim = 0. We say that x→∞ g(x) f (x) f (x) is aymptotic to g(x) and write f (x) ∼ g(x) if lim = 1. Analogous definitions x→∞ g(x) exist if the domain is in the set of positive integers. • Examples. Discuss each of the following: n n k 2 k∼ n , k=1 k=1 √ n2 , 2 x+1− √ x 1 √ , x log 1 + 1 x = O(1/x). • Comment: The expression O(g(x)) in an equation represents a function f (x) = O(g(x)). To clarify, the last equation in p≤x x = p p≤x x +O p does not assert that a function is O 1 p≤x 1 = p≤x x + O(x) p if and only if it is O(x) but rather there is a p≤x function f (x) that satisfies f (x) = O 1 and f (x) = O(x).

T2 More precisely, the analysis above gives n≤x 1 = log x + γ + E(x) n where E(x) = − {x} + x ∞ x {t} dt. t2 Recall that this last integral is ≤ 1/x so that we can deduce |E(x)| ≤ 1/x. 392726 . . n≤106 44 • Comment: The constant γ is called Euler’s constant. It is unknown whether or not γ is irrational. Sums and Products of Primes Revisited: • We combine the lemma from the previous section with the Prime Number Theorem to arrive at improvements to some earlier results. Theorem 37: There exist constants C1 , C2 , and C3 such that 1 1 (i) = log log x + C1 + O , p log x p≤x (ii) p≤x 1 log p , and = log x + C2 + O p log x 1− (iii) p≤x 1 p C3 .

P≤z Taking z = log x, we obtain 1− A1 (z, x) = x p≤log x 1 p2 + O 2log x = x p≤log x Thus, A1 (z, x) ∼ (6/π 2 )x (with z = log x). Since the number of squarefree numbers ≤ x is A1 (z, x) − A2 (z, x), it suffices to show A2 (z, x) = o(x). We use that A2 (z, x) ≤ 1= n≤x p>z p2 |n 1= p>z n≤x p2 |n p>z 1 . p2 ∞ 1 1 . Since z = log x and n2 p2 p p>z n=1 1 is the tail end of a convergent series, we deduce that = o(1). It follows that p2 p>z The series 1 converges by comparison with p2 p>z x ≤x p2 A2 (z, x) = o(x), completing the proof.