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Question Number 28619 by abdo imad last updated on 27/Jan/18
calculate Σ_(k=2) ^(+∞) ln(1−(1/k^2 )) .
$${calculate}\:\:\sum_{{k}=\mathrm{2}} ^{+\infty} \:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\:. \\ $$
Commented by abdo imad last updated on 31/Jan/18
let put S_n = Σ_(k=2) ^n ln(1−(1/k^2 ) ) and S=Σ_(k=2) ^( ∞) ln(1−(1/k^2 )) we have S=lim_(n→+∞ ) S_n but S_n =ln(Π_(k=2) ^n (1−(1/k^2 ))) but Π_(k=2) ^n (1−(1/k^2 ))= Π_(k=2) ^n (((k^2 −1)/k^2 ))⇒ln(Π_(k=2) ^n (1−(1/k^2 ))) =Σ_(k=2) ^n ln(k−1) +Σ_(k=2) ^n ln(k+1) −2Σ_2 ^n ln(k) = Σ_(k=1) ^(n−1) ln(k) +Σ_(k=3) ^(n+1) ln(k) −2Σ_(k=2) ^n ln(k) =Σ_(k=2) ^n lnk −ln(n) +Σ_(k=2) ^n ln(k)−ln2+ln(n+1)−2Σ_(k=2) ^n ln(k) =−ln(n)+ln(n+1) −ln2=ln(((n+1)/n)) −ln(2) so lim_(n→+∞) S_n = −ln(2) = S .
$${let}\:{put}\:{S}_{{n}} =\:\sum_{{k}=\mathrm{2}} ^{{n}} \:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:\right)\:{and}\:{S}=\sum_{{k}=\mathrm{2}} ^{\:\infty} \:{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right) \\ $$$${we}\:{have}\:{S}={lim}_{{n}\rightarrow+\infty\:} \:{S}_{{n}} \:\:{but} \\ $$$${S}_{{n}} ={ln}\left(\prod_{{k}=\mathrm{2}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\right)\:\:{but} \\ $$$$\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)=\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\frac{{k}^{\mathrm{2}} −\mathrm{1}}{{k}^{\mathrm{2}} }\right)\Rightarrow{ln}\left(\prod_{{k}=\mathrm{2}} ^{{n}} \left(\mathrm{1}−\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\right)\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left({k}−\mathrm{1}\right)\:+\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left({k}+\mathrm{1}\right)\:−\mathrm{2}\sum_{\mathrm{2}} ^{{n}} \:{ln}\left({k}\right) \\ $$$$=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {ln}\left({k}\right)\:+\sum_{{k}=\mathrm{3}} ^{{n}+\mathrm{1}} {ln}\left({k}\right)\:−\mathrm{2}\sum_{{k}=\mathrm{2}} ^{{n}} \:{ln}\left({k}\right) \\ $$$$=\sum_{{k}=\mathrm{2}} ^{{n}} {lnk}\:−{ln}\left({n}\right)\:+\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left({k}\right)−{ln}\mathrm{2}+{ln}\left({n}+\mathrm{1}\right)−\mathrm{2}\sum_{{k}=\mathrm{2}} ^{{n}} {ln}\left({k}\right) \\ $$$$=−{ln}\left({n}\right)+{ln}\left({n}+\mathrm{1}\right)\:−{ln}\mathrm{2}={ln}\left(\frac{{n}+\mathrm{1}}{{n}}\right)\:−{ln}\left(\mathrm{2}\right) \\ $$$${so}\:{lim}_{{n}\rightarrow+\infty} \:{S}_{{n}} =\:−{ln}\left(\mathrm{2}\right)\:\:=\:{S}\:\:. \\ $$$$ \\ $$

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