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Question Number 156413 by KONE last updated on 10/Oct/21
Answered by KONE last updated on 10/Oct/21
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Answered by mindispower last updated on 11/Oct/21
(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
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