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Question Number 106378 by mr W last updated on 04/Aug/20
Answered by Dwaipayan Shikari last updated on 05/Aug/20
(1−12)(1−14)(1−16)....n∏nn=1(1−12n)=ylog(∏nn=1(1−12n))=logy∑nn=1log(1−12n)=logy∑nn=1log(1−12n)n⩾(∏nn=1log(1−12n))1nlog(12.34......2n−12n)n⩾(∏nlog(1−12n))1n
Commented by mr W last updated on 05/Aug/20
thanks,butitdoesn′tanswertheactualquestion.
Answered by mr W last updated on 05/Aug/20
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