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Question Number 52305 by Abdo msup. last updated on 05/Jan/19

calculate Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1)))

calculate∑n=1∞(−1)nn(4n2−1)

Commented by Abdo msup. last updated on 06/Jan/19

let decompose F(x)=(1/(x(4x^2 −1)))=(1/(x(2x+1)(2x−1))) ⇒ F(x)=(a/x) +(b/(2x+1)) +(c/(2x−1)) a=lim_(x→0) xF(x)=−1 b=lim_(x→−(1/2)) (2x+1)F(x) =(1/((−(1/2))(−2))) =1 c=lim_(x→(1/2)) (2x−1)F(x)=1 ⇒ F(x)=−(1/x) +(1/(2x+1)) +(1/(2x−1)) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1))) =−Σ_(n=1) ^∞ (((−1)^n )/n) +Σ_(n=1) ^∞ (((−1)^n )/(2n+1)) +Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) but 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 Σ_(n=1) ^∞ (((−1)^n )/(2n−1)) =Σ_(n=0) ^∞ (((−1)^(n−1) )/(2n+1)) =−(π/4) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1))) =ln(2)+(π/4) −1−(π/4) =ln(2)−1 .

letdecomposeF(x)=1x(4x2−1)=1x(2x+1)(2x−1)⇒F(x)=ax+b2x+1+c2x−1a=limx→0xF(x)=−1b=limx→−12(2x+1)F(x)=1(−12)(−2)=1c=limx→12(2x−1)F(x)=1⇒F(x)=−1x+12x+1+12x−1⇒∑n=1∞(−1)nn(4n2−1)=−∑n=1∞(−1)nn+∑n=1∞(−1)n2n+1+∑n=1∞(−1)n2n−1but∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)n2n+1=∑n=0∞(−1)n2n+1−1=π4−1∑n=1∞(−1)n2n−1=∑n=0∞(−1)n−12n+1=−π4⇒∑n=1∞(−1)nn(4n2−1)=ln(2)+π4−1−π4=ln(2)−1.

Answered by Smail last updated on 06/Jan/19

Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1)))=Σ_(n=1) ^∞ (((−1)^n )/(n(2n+1)(2n−1))) (1/(n(2n+1)(2n−1)))=(a/n)+(b/(2n+1))+(c/(2n−1)) a=−1 , b=1 , c=1 Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1)))=Σ_(n=1) ^∞ (((−1)^n )/(2n+1))+Σ_(n=1) ^∞ (((−1)^n )/(2n−1))−Σ_(n=1) ^∞ (((−1)^n )/n) let put p(x)=Σ_(n=1) ^∞ (((−1)^n )/(2n+1))x^(2n+1) +Σ_(n=1) ^∞ (((−1)^n )/(2n−1))x^(2n−1) +Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n =Σ_(n=1) ^∞ (((−1)^n )/(2n+1))x^(2n+1) +Σ_(n=0) ^∞ (((−1)^(n−1) )/(2n+1))x^(2n+1) +Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n =Σ_(n=1) ^∞ (((−1)^n )/(2n+1))x^(2n+1) −Σ_(n=0) ^∞ (((−1)^n )/(2n+1))x^(2n+1) +Σ_(n=1) ^∞ (((−1)^(n−1) )/n)x^n =−x+ln(1+x) with ∣x∣≤1 p(1)=ln(2)−1 Which means Σ_(n=1) ^∞ (((−1)^n )/(n(4n^2 −1)))=ln(2)−1

∑∞n=1(−1)nn(4n2−1)=∑∞n=1(−1)nn(2n+1)(2n−1)1n(2n+1)(2n−1)=an+b2n+1+c2n−1a=−1,b=1,c=1∑∞n=1(−1)nn(4n2−1)=∑∞n=1(−1)n2n+1+∑∞n=1(−1)n2n−1−∑∞n=1(−1)nnletputp(x)=∑∞n=1(−1)n2n+1x2n+1+∑∞n=1(−1)n2n−1x2n−1+∑∞n=1(−1)n−1nxn=∑∞n=1(−1)n2n+1x2n+1+∑∞n=0(−1)n−12n+1x2n+1+∑∞n=1(−1)n−1nxn=∑∞n=1(−1)n2n+1x2n+1−∑∞n=0(−1)n2n+1x2n+1+∑∞n=1(−1)n−1nxn=−x+ln(1+x)with∣x∣⩽1p(1)=ln(2)−1Whichmeans∑∞n=1(−1)nn(4n2−1)=ln(2)−1

Commented by Abdo msup. last updated on 06/Jan/19

thanks sir Smail.

thankssirSmail.

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