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Question Number 1763 by Gerlândio Almeida last updated on 18/Sep/15


Commented by 123456 last updated on 18/Sep/15
$p_0 (n)=xn r(n)=αxn p_1 (n)=p_0 (n)−r(n)=(1−α)xn r(n)≥p_0 (m) αxn≥xm x>0 n≥(m/α) α>0 p_0 (n)≥((xm)/α) r_0 (n)≥xm p_1 (n)≥(((1−α)xm)/α) (n,m)∈Z^2 α\m 1 2 3 4 5 0,20 5 10 15 20 25 0,35 3 6 9 12 15 0,50 2 4 6 8 10$
$p_{0} (n) = x n$ $r (n) = α x n$ $p_{1} (n) = p_{0} (n) - r (n) = (1 - α) x n$ $r (n) ⩾ p_{0} (m)$ $α x n ⩾ x m x > 0$ $n ⩾ \frac{m}{α} α > 0$ $p_{0} (n) ⩾ \frac{x m}{α}$ $r_{0} (n) ⩾ x m$ $p_{1} (n) ⩾ \frac{(1 - α) x m}{α}$ $(n, m) \in Z^{2}$ $α ∖ m 1 2 3 4 5$ $0, 20 5 10 15 20 25$ $0, 35 3 6 9 12 15$ $0, 50 2 4 6 8 10$
	Answered by 123456 last updated on 18/Sep/15

	$\underset{n = 0}{\sum^{\infty}} x^{n} = \frac{1}{1 - x}$ $\prod_{p \in P} \frac{1}{1 - p^{- s}} = \prod_{p \in P} \underset{n = 0}{\sum^{\infty}} p^{- s n} = \underset{n = 1}{\sum^{\infty}} n^{- s}, s > 1$
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