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

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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

$${p}_{\mathrm{0}} \left({n}\right)={xn} \\ $$$${r}\left({n}\right)=\alpha{xn} \\ $$$${p}_{\mathrm{1}} \left({n}\right)={p}_{\mathrm{0}} \left({n}\right)−{r}\left({n}\right)=\left(\mathrm{1}−\alpha\right){xn} \\ $$$${r}\left({n}\right)\geqslant{p}_{\mathrm{0}} \left({m}\right) \\ $$$$\alpha{xn}\geqslant{xm}\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{0} \\ $$$${n}\geqslant\frac{{m}}{\alpha}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha>\mathrm{0} \\ $$$${p}_{\mathrm{0}} \left({n}\right)\geqslant\frac{{xm}}{\alpha} \\ $$$${r}_{\mathrm{0}} \left({n}\right)\geqslant{xm} \\ $$$${p}_{\mathrm{1}} \left({n}\right)\geqslant\frac{\left(\mathrm{1}−\alpha\right){xm}}{\alpha} \\ $$$$\left({n},{m}\right)\in\mathbb{Z}^{\mathrm{2}} \\ $$$$\alpha\backslash{m}\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{5} \\ $$$$\mathrm{0},\mathrm{20}\:\:\:\:\mathrm{5}\:\:\:\:\:\mathrm{10}\:\:\:\:\mathrm{15}\:\:\:\:\:\:\mathrm{20}\:\:\:\:\:\:\mathrm{25} \\ $$$$\mathrm{0},\mathrm{35}\:\:\:\:\mathrm{3}\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\mathrm{9}\:\:\:\:\:\:\:\mathrm{12}\:\:\:\:\:\:\mathrm{15} \\ $$$$\mathrm{0},\mathrm{50}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\mathrm{6}\:\:\:\:\:\:\:\:\mathrm{8}\:\:\:\:\:\:\:\:\mathrm{10} \\ $$

Answered by 123456 last updated on 18/Sep/15

Σ_(n=0) ^∞ x^n =(1/(1−x)) Π_(p∈P) (1/(1−p^(−s) ))=Π_(p∈P) Σ_(n=0) ^∞ p^(−sn) =Σ_(n=1) ^∞ n^(−s) ,s>1

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\underset{{p}\in\mathbb{P}} {\prod}\frac{\mathrm{1}}{\mathrm{1}−{p}^{−{s}} }=\underset{{p}\in\mathbb{P}} {\prod}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{p}^{−{sn}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{n}^{−{s}} ,{s}>\mathrm{1} \\ $$

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