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Question Number 68593 by Abdo msup. last updated on 14/Sep/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)^2 ))
calculate∑n=1∞(−1)nn(2n+1)2
Commented by Abdo msup. last updated on 06/Oct/19
let S=Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)^2 )) first let decompose F(x)=(1/(x(2x+1)^2 )) ⇒F(x)=(a/x) +(b/(2x+1)) +(c/((2x+1)^2 )) a=lim_(x→0) xF(x)=1 c=lim_(x→−(1/2)) (2x+1)^2 F−x)=−2 ⇒ F(x)=(1/x) +(b/(2x+1))−(2/((2x+1)^2 )) lim_(x→+∞) xF(x)=0 =1+(b/2) ⇒b+2=0 ⇒b=−2 ⇒ F(x)=(1/x) −(2/(2x+1)) −(2/((2x+1)^2 )) ⇒ S=Σ_(n=1) ^∞ (((−1)^n )/n) −2Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) −2Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) we have Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/((2n+1))) =Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) −1=(π/4)−1 rest to ca<culate α_0 =Σ_(n=1) ^∞ (((−1)^n )/((2n+1)^2 )) ⇒ S=−ln(2)−(π/2) +2 −2α_0
letS=∑n=1∞(−1)nn(2n+1)2firstletdecomposeF(x)=1x(2x+1)2⇒F(x)=ax+b2x+1+c(2x+1)2a=limx→0xF(x)=1c=limx→−12(2x+1)2F−x)=−2⇒F(x)=1x+b2x+1−2(2x+1)2limx→+∞xF(x)=0=1+b2⇒b+2=0⇒b=−2⇒F(x)=1x−22x+1−2(2x+1)2⇒S=∑n=1∞(−1)nn−2∑n=1∞(−1)n2n+1−2∑n=1∞(−1)n(2n+1)2wehave∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)n(2n+1)=∑n=0∞(−1)n2n+1−1=π4−1resttoca<culateα0=∑n=1∞(−1)n(2n+1)2⇒S=−ln(2)−π2+2−2α0

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