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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 2

svpbesoindaidepourlaquestion2

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)

(3−1n+1)⩾(3−1n)(1+1(n+1)3)...(1)lemma(1)⇔1n−1n+1⩾3(n+1)3−1n(n+1)3(n+1)3−n(n+1)2⩾3n−1(n+1)2n⩾3n−1truesince(n+1)2⩾4n(n+1)2⩾4n=3n+n⩾3n−1.....(1)TrueletsprouveThisn=1wehave2⩽3−1=2truesuppose∀n∈N∗∏k⩽n(1+1k3)⩽3−1nletprouv∏k⩽n+1(1+1k3)⩽3−1n+1∏0⩽k⩽n+1(1+1k3)=∏0⩽k⩽n(1+1k3).(1+1(n+1)3)⩽(3−1n)(1+1(n+1)3)ByHypothesis⩽3−1n+1UsingLemma(1)⇒∀n∈N∗∏0⩽k⩽n(1+1k3)⩽3−1n

Commented by KONE last updated on 13/Oct/21

merci bien a vous

mercibienavous

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