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Question Number 34688 by math khazana by abdo last updated on 09/May/18
$${cslculate}\:\sum_{{n}=\mathrm{2}} ^{\infty} \:{ln}\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\right) \\ $$
Commented by abdo mathsup 649 cc last updated on 14/May/18
![let put S_n = Σ_(k=2) ^n ln(1+(((−1)^k )/k)) S_n = ln{ Π_(k=2) ^n (1+(((−1)^k )/k))}=ln(w_n ) but w_n = Π_(k=2) ^n (1+(((−1)^k )/k)) = Π_(p=1) ^([(n/2)]) ( 1+ (1/(2p))) Π_(p=1) ^([((n−1)/2)]) (1−(1/(2p+1))) w_(2n) =Π_(p=1) ^n ((2p+1)/(2p)) Π_(p=1) ^(n−1) ((2p)/(2p+1)) = ((2n−1)/(2n−2)) Π_(p=1) ^(n−1) (1) = ((2n−1)/(2n−2)) →1(n→+∞) lim_(n→+∞) S_(2n) = 0 w_(2n+1) = Π_(p=1) ^n (1−(1/(2p+1)))Π_(p=1) ^n (1 +(1/(2p))) = Π_(p=1) ^n ((2p)/(2p+1)) Π_(p=1) ^n ((2p+1)/(2p)) =1 ⇒lim S_(2n+1) =0 so S_n →0(n→+∞)](Q34981.png)
$${let}\:{put}\:{S}_{{n}} \:=\:\sum_{{k}=\mathrm{2}} ^{{n}} \:{ln}\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\right) \\ $$$${S}_{{n}} =\:{ln}\left\{\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\right)\right\}={ln}\left({w}_{{n}} \right)\:\:{but} \\ $$$${w}_{{n}} =\:\prod_{{k}=\mathrm{2}} ^{{n}} \:\left(\mathrm{1}+\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\right)\:=\:\prod_{{p}=\mathrm{1}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}{p}}\right)\:\prod_{{p}=\mathrm{1}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\right) \\ $$$${w}_{\mathrm{2}{n}} \:=\prod_{{p}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}{p}}\:\prod_{{p}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\:\frac{\mathrm{2}{p}}{\mathrm{2}{p}+\mathrm{1}} \\ $$$$=\:\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}−\mathrm{2}}\:\prod_{{p}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\left(\mathrm{1}\right)\:=\:\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}−\mathrm{2}}\:\rightarrow\mathrm{1}\left({n}\rightarrow+\infty\right) \\ $$$${lim}_{{n}\rightarrow+\infty} \:\:{S}_{\mathrm{2}{n}} \:\:=\:\mathrm{0} \\ $$$${w}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\prod_{{p}=\mathrm{1}} ^{{n}} \:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{1}}\right)\prod_{{p}=\mathrm{1}} ^{{n}} \:\left(\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}{p}}\right) \\ $$$$=\:\prod_{{p}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{2}{p}}{\mathrm{2}{p}+\mathrm{1}}\:\prod_{{p}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{2}{p}+\mathrm{1}}{\mathrm{2}{p}}\:=\mathrm{1}\:\Rightarrow{lim}\:{S}_{\mathrm{2}{n}+\mathrm{1}} =\mathrm{0}\:{so} \\ $$$${S}_{{n}} \:\rightarrow\mathrm{0}\left({n}\rightarrow+\infty\right) \\ $$$$ \\ $$