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Question Number 57529 by maxmathsup by imad last updated on 06/Apr/19
calculate Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)))
calculate∑n=1∞(−1)nn3(n+1)
Commented by maxmathsup by imad last updated on 09/Apr/19
let decompose F(x)=(1/((x+1)x^3 )) ⇒F(x)=(a/(x+1)) +(b/x) +(c/x^2 ) +(d/x^3 ) a =lim_(x→−1) (x+1)F(x)=−1 d =lim_(x→0) x^3 F(x) =1 ⇒F(x) =((−1)/(x+1)) +(b/x) +(c/x^2 ) +(1/x^3 ) F(1) =(1/2) =−(1/2) +b+c +1 ⇒b+c =0 ⇒c=−b F(−2) =(1/8) =1 −(b/2) +(c/4)−(1/8) ⇒1 =8−4b +2c−1 ⇒−6 =−4b +2c ⇒−3 =−2b +c b =1 and c =−1 ⇒F(x) =−(1/(x+1)) +(1/x) −(1/x^2 ) +(1/x^3 ) ⇒ Σ_(n=1) ^∞ (((−1)^n )/((n+1)n^3 )) =−Σ_(n=1) ^∞ (((−1)^n )/(n+1)) +Σ_(n=1) ^∞ (((−1)^n )/n) − Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +Σ_(n=1) ^∞ (((−1)^n )/n^3 ) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n) −1 =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) let δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) let determine δ(x) interms of ξ(x) (x>1) we have δ(x) =Σ_(n=1) ^∞ (1/(2^x n^x )) −Σ_(n=0) ^∞ (1/((2n+1)^x )) and Σ_(n=1) ^∞ (1/n^x ) =Σ_(n=1) ^∞ (1/(2^x n^x )) +Σ_(n=0) ^∞ (1/((2n+1)^x )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^x )) =ξ(x)−2^(−x) ξ(x) =(1−2^(−x) )ξ(x) ⇒δ(x) =2^(−x) ξ(x)−(1−2^(−x) )ξ(x)=(2^(−x) −1−2^(−x) )ξ(x) =(2^(1−x) −1)ξ(x) ⇒ δ(x)=(2^(1−x) −1)ξ(x) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =δ(2) =(2^(1−2) −1)ξ(2) =((1/2)−1)(π^2 /6) =−(π^2 /(12)) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^3 ) =δ(3) =(2^(1−3) −1)ξ(3) =((1/4)−1)ξ(3)=−(3/4)ξ(3) Σ_(n=1) ^∞ (((−1)^n )/((n+1)n^3 )) =1−ln(2) −ln(2) −(−(π^2 /(12)))−(3/4)ξ(3) =1−2ln(2)+(π^2 /(12)) −(3/4) ξ(3) with ξ(3) =Σ_(n=1) ^∞ (1/n^3 ) .
letdecomposeF(x)=1(x+1)x3⇒F(x)=ax+1+bx+cx2+dx3a=limx→−1(x+1)F(x)=−1d=limx→0x3F(x)=1⇒F(x)=−1x+1+bx+cx2+1x3F(1)=12=−12+b+c+1⇒b+c=0⇒c=−bF(−2)=18=1−b2+c4−18⇒1=8−4b+2c−1⇒−6=−4b+2c⇒−3=−2b+cb=1andc=−1⇒F(x)=−1x+1+1x−1x2+1x3⇒∑n=1∞(−1)n(n+1)n3=−∑n=1∞(−1)nn+1+∑n=1∞(−1)nn−∑n=1∞(−1)nn2+∑n=1∞(−1)nn3∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=∑n=1∞(−1)n−1n−1=ln(2)−1∑n=1∞(−1)nn=−ln(2)letδ(x)=∑n=1∞(−1)nnxletdetermineδ(x)intermsofξ(x)(x>1)wehaveδ(x)=∑n=1∞12xnx−∑n=0∞1(2n+1)xand∑n=1∞1nx=∑n=1∞12xnx+∑n=0∞1(2n+1)x⇒∑n=0∞1(2n+1)x=ξ(x)−2−xξ(x)=(1−2−x)ξ(x)⇒δ(x)=2−xξ(x)−(1−2−x)ξ(x)=(2−x−1−2−x)ξ(x)=(21−x−1)ξ(x)⇒δ(x)=(21−x−1)ξ(x)⇒∑n=1∞(−1)nn2=δ(2)=(21−2−1)ξ(2)=(12−1)π26=−π212⇒∑n=1∞(−1)nn3=δ(3)=(21−3−1)ξ(3)=(14−1)ξ(3)=−34ξ(3)∑n=1∞(−1)n(n+1)n3=1−ln(2)−ln(2)−(−π212)−34ξ(3)=1−2ln(2)+π212−34ξ(3)withξ(3)=∑n=1∞1n3.

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