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
$${let}\:{put}\:\:{A}_{{n}} =\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\:\:\mathrm{1}\:−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$$=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{\left({k}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)}{{k}^{\mathrm{2}} }\:=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}\:.\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}+\mathrm{1}}{{k}}\:{but} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\frac{{k}−\mathrm{1}}{{k}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{4}}.....\frac{{n}−\mathrm{1}}{{n}}\:=\:\frac{\mathrm{1}}{{n}} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\:\:\frac{{k}+\mathrm{1}}{{k}}\:=\:\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{5}}{\mathrm{4}}....\frac{{n}}{{n}−\mathrm{1}}.\frac{{n}+\mathrm{1}}{{n}}\:=\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}_{{n}} =\:\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\:\:{and}\:{lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Commented by abdo mathsup 649 cc last updated on 29/May/18
$${remark}\:{we}\:{have}\:{ln}\left({A}_{{n}} \right)=\:\sum_{{k}=\mathrm{2}} ^{{n}} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$${ln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)\:\Rightarrow{lim}_{{n}\rightarrow+\infty} {ln}\left({A}_{{n}} \right)\:=−{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{2}} ^{\infty} \:\:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:=−{ln}\left(\mathrm{2}\right) \\ $$
Commented by bshahid010@gmail.com last updated on 30/May/18
thanks