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Question Number 36163 by bshahid010@gmail.com last updated on 29/May/18

Commented by abdo mathsup 649 cc last updated on 29/May/18

let put A_n = Π_(k=2) ^n ( 1 −(1/k^2 )) = Π_(k=2) ^n (((k−1)(k+1))/k^2 ) = Π_(k=2) ^n ((k−1)/k) .Π_(k=2) ^n ((k+1)/k) but Π_(k=2) ^n ((k−1)/k) = (1/2).(2/3).(3/4).....((n−1)/n) = (1/n) Π_(k=2) ^n ((k+1)/k) = (3/2).(4/3).(5/4)....(n/(n−1)).((n+1)/n) = ((n+1)/2) ⇒ A_n = ((n+1)/(2n)) and lim_(n→+∞) A_n = (1/2) .

letputAn=∏k=2n(1−1k2)=∏k=2n(k−1)(k+1)k2=∏k=2nk−1k.∏k=2nk+1kbut∏k=2nk−1k=12.23.34.....n−1n=1n∏k=2nk+1k=32.43.54....nn−1.n+1n=n+12⇒An=n+12nandlimn→+∞An=12.

Commented by abdo mathsup 649 cc last updated on 29/May/18

remark we have ln(A_n )= Σ_(k=2) ^n ln(1−(1/k^2 )) ln(((n+1)/(2n))) ⇒lim_(n→+∞) ln(A_n ) =−ln(2) ⇒ Σ_(n=2) ^∞ ln(1−(1/n^2 )) =−ln(2)

remarkwehaveln(An)=∑k=2nln(1−1k2)ln(n+12n)⇒limn→+∞ln(An)=−ln(2)⇒∑n=2∞ln(1−1n2)=−ln(2)

Commented by bshahid010@gmail.com last updated on 30/May/18

thanks

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