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Question Number 33305 by abdo imad last updated on 14/Apr/18

find Σ_(n=2) ^∞  (1 −(1/n^2 ))

$${find}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:\left(\mathrm{1}\:−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$

Commented by prof Abdo imad last updated on 15/Apr/18

the Q is find Σ_(n=2) ^∞  ln(1−(1/n^2 ))

$${the}\:{Q}\:{is}\:{find}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right) \\ $$

Commented by abdo imad last updated on 19/Apr/18

let put S_n = Σ_(k=2) ^n ln(1−(1/k^2 ))  we have S_n =ln(Π_(k=2) ^n  (1−(1/k^2 ))) but we have  Π_(k=2) ^n  (1−(1/k^2 )) = Π_(k=2) ^n  ((k^2 −1)/k^2 ) =Π_(k=2) ^n  ((k−1)/k) .((k+1)/k)  =((1/2) (2/3) (3/4) ....((n−2)/(n−1)) ((n−1)/n))((3/2) (4/3) (5/4) ....(n/(n−1)) ((n+1)/n))  =(1/n) ((n+1)/2) =((n+1)/(2n)) ⇒ S_n  =ln(((n+1)/(2n)))⇒lim_(n→+∞ ) S_n =−ln(2)  so Σ_(n=2) ^∞ ln(1−(1/n^2 ))=−ln(2) .

$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$${we}\:{have}\:{S}_{{n}} ={ln}\left(\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\right)\:{but}\:{we}\:{have} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\:=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\:=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{k}−\mathrm{1}}{{k}}\:.\frac{{k}+\mathrm{1}}{{k}} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{3}}{\mathrm{4}}\:....\frac{{n}−\mathrm{2}}{{n}−\mathrm{1}}\:\frac{{n}−\mathrm{1}}{{n}}\right)\left(\frac{\mathrm{3}}{\mathrm{2}}\:\frac{\mathrm{4}}{\mathrm{3}}\:\frac{\mathrm{5}}{\mathrm{4}}\:....\frac{{n}}{{n}−\mathrm{1}}\:\frac{{n}+\mathrm{1}}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{{n}}\:\frac{{n}+\mathrm{1}}{\mathrm{2}}\:=\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\:\Rightarrow\:{S}_{{n}} \:={ln}\left(\frac{{n}+\mathrm{1}}{\mathrm{2}{n}}\right)\Rightarrow{lim}_{{n}\rightarrow+\infty\:} {S}_{{n}} =−{ln}\left(\mathrm{2}\right) \\ $$$${so}\:\sum_{{n}=\mathrm{2}} ^{\infty} {ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)=−{ln}\left(\mathrm{2}\right)\:. \\ $$

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