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Question-36163




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) .
$${let}\:{put}\:\:{A}_{{n}} =\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\:\:\mathrm{1}\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$$=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{{k}^{\mathrm{2}} }\:=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}\:.\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}+\mathrm{1}}{{k}}\:{but} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{{k}−\mathrm{1}}{{k}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}…..\frac{{n}−\mathrm{1}}{{n}}\:=\:\frac{\mathrm{1}}{{n}} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{{k}+\mathrm{1}}{{k}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{5}}{\mathrm{4}}….\frac{{n}}{{n}−\mathrm{1}}.\frac{{n}+\mathrm{1}}{{n}}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}_{{n}} =\:\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$ \\ $$
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)
$${remark}\:{we}\:{have}\:{ln}\left({A}_{{n}} \right)=\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$${ln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {ln}\left({A}_{{n}} \right)\:=−{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:=−{ln}\left(\mathrm{2}\right) \\ $$
Commented by bshahid010@gmail.com last updated on 30/May/18
thanks

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