Question-156413

Question Number 156413 by KONE last updated on 10/Oct/21

Answered by KONE last updated on 10/Oct/21

$${svp}\:{besoin}\:{d}\:{aide}\:{pour}\:{la}\:{question}\:\mathrm{2} \\ $$

Answered by mindispower last updated on 11/Oct/21

(3−(1/(n+1)))≥(3−(1/n))(1+(1/((n+1)^3 )))...(1) lemma(1) ⇔(1/n)−(1/(n+1))≥(3/((n+1)^3 ))−(1/(n(n+1)^3 )) (n+1)^3 −n(n+1)^2 ≥3n−1 (n+1)^2 n≥3n−1 true since (n+1)^2 ≥4 n(n+1)^2 ≥4n=3n+n≥3n−1..... (1) True lets prouve This n=1 we have 2≤3−1=2 true suppose ∀n∈N^∗ Π_(k≤n) (1+(1/k^3 ))≤3−(1/n) let prouv Π_(k≤n+1) (1+(1/k^3 ))≤3−(1/(n+1)) Π_(0≤k≤n+1) (1+(1/k^3 ))=Π_(0≤k≤n) (1+(1/k^3 )).(1+(1/((n+1)^3 )))≤ (3−(1/n))(1+(1/((n+1)^3 )))_(By Hypothesis) ≤3−(1/(n+1)) Using Lemma (1) ⇒∀n∈N^∗ Π_(0≤k≤n) (1+(1/k^3 ))≤3−(1/n)

$$\left(\mathrm{3}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right)\geqslant\left(\mathrm{3}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right)…\left(\mathrm{1}\right) \\ $$$${lemma}\left(\mathrm{1}\right) \\ $$$$\Leftrightarrow\frac{\mathrm{1}}{{n}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\geqslant\frac{\mathrm{3}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{3}} −{n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{3}{n}−\mathrm{1} \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{2}} {n}\geqslant\mathrm{3}{n}−\mathrm{1} \\ $$$${true}\:{since}\: \\ $$$$\left({n}+\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{4} \\ $$$${n}\left({n}+\mathrm{1}\right)^{\mathrm{2}} \geqslant\mathrm{4}{n}=\mathrm{3}{n}+{n}\geqslant\mathrm{3}{n}−\mathrm{1}….. \\ $$$$\left(\mathrm{1}\right)\:{True} \\ $$$${lets}\:{prouve}\:{This} \\ $$$${n}=\mathrm{1} \\ $$$${we}\:{have}\:\mathrm{2}\leqslant\mathrm{3}−\mathrm{1}=\mathrm{2}\:{true} \\ $$$${suppose}\:\forall{n}\in\mathbb{N}^{\ast} \:\underset{{k}\leqslant{n}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)\leqslant\mathrm{3}−\frac{\mathrm{1}}{{n}} \\ $$$${let}\:{prouv}\:\underset{{k}\leqslant{n}+\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)\leqslant\mathrm{3}−\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$\underset{\mathrm{0}\leqslant{k}\leqslant{n}+\mathrm{1}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)=\underset{\mathrm{0}\leqslant{k}\leqslant{n}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right).\left(\mathrm{1}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right)\leqslant \\ $$$$\left(\mathrm{3}−\frac{\mathrm{1}}{{n}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} }\right)_{{By}\:{Hypothesis}} \leqslant\mathrm{3}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\:\:{Using}\:{Lemma}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow\forall{n}\in\mathbb{N}^{\ast} \underset{\mathrm{0}\leqslant{k}\leqslant{n}} {\prod}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}^{\mathrm{3}} }\right)\leqslant\mathrm{3}−\frac{\mathrm{1}}{{n}} \\ $$$$ \\ $$

Commented by KONE last updated on 13/Oct/21

$${merci}\:{bien}\:{a}\:{vous} \\ $$

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