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Question Number 60499 by abdo mathsup 649 cc last updated on 21/May/19
find the value of Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^4 ))
findthevalueof∑n=1∞(−1)nn3(n+1)4
Commented by maxmathsup by imad last updated on 30/May/19
let decompose F(x) =(1/(x^3 (x+1)^4 )) we have F(x) =Σ_(i=1) ^3 (a_i /x^i ) +Σ_(i=1) ^4 (b_i /((x+1)^i )) let determine D_2 (0) for f(x) =(x+1)^(−4) we have f(x)=f(0) +((f^′ (0))/(1!)) x +((f^((2)) (0))/(2!)) x^2 +(x^3 /(3!))ξ(x) f(0) =1 ,f^′ (x) =−4(x+1)^(−5) ⇒f^′ (0) =−4 ,f^((2)) (x)=20 (x+1)^(−6) ⇒ f^((2)) (0) =20 ⇒f(x)=1−4x +10x^2 +x^3 ξ(x) ⇒ ((f(x))/x^3 ) =(1/x^3 ) −(4/x^2 ) +((10)/x) +ξ(x) ⇒a_1 =10 , a_2 =−4 , a_3 =−1 also we have F(x) =_(x+1=t) (1/((t−1)^3 t^4 )) let find D_3 (0) for g(t) =(t−1)^(−3) g(t) =g(0) +((g^′ (0))/(1!)) t +((g^((2)) (0))/(2!)) t^2 +((g^((3)) (0))/(3!))t^3 +(t^4 /(4!))ξ(t) g(0) =−1 , g^′ (t) =−3(t−1)^(−4) ⇒g^′ (0) =−3 g^((2)) (t) =12(t−1)^(−5) ⇒g^((2)) (0) =−12 , g^((3)) (t) =−60(t−1)^(−6) ⇒ g^((3)) (0) =−60 ⇒ g(t) =−1 −3t −6t^2 −10 t^3 +(t^4 /(4!))ξ(t) ⇒ ((g(t))/t^4 ) =−(1/t^4 ) −(3/t^3 ) −(6/t^2 ) −((10)/t) +(1/(4!))ξ(t) =−(1/((x+1)^4 )) −(3/((x+1)^3 )) −(6/((x+1)^2 )) −((10)/((x+1))) +(1/(4!))ξ(x+1) ⇒ b_1 =−10 , b_2 =−6 , b_3 =−3 , b_4 =−1 and F(x) =((10)/x) −(4/x^2 ) −(1/x^3 ) −((10)/((x+1))) −(6/((x+1)^2 )) −(3/((x+1)^3 )) −(1/((x+1)^4 )) ⇒ Σ_(n=1) ^∞ (((−1)^n )/(n^3 (n+1)^4 )) =Σ_(n=1) ^∞ (−1)^n F(n) =10 Σ_(n=1) ^∞ (((−1)^n )/n) −4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) −10 Σ_(n=1) ^∞ (((−1)^n )/(n+1)) −6Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) −3 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) −Σ_(n=1) ^∞ (((−1)^n )/n^4 )
letdecomposeF(x)=1x3(x+1)4wehaveF(x)=∑i=13aixi+∑i=14bi(x+1)iletdetermineD2(0)forf(x)=(x+1)−4wehavef(x)=f(0)+f′(0)1!x+f(2)(0)2!x2+x33!ξ(x)f(0)=1,f′(x)=−4(x+1)−5⇒f′(0)=−4,f(2)(x)=20(x+1)−6⇒f(2)(0)=20⇒f(x)=1−4x+10x2+x3ξ(x)⇒f(x)x3=1x3−4x2+10x+ξ(x)⇒a1=10,a2=−4,a3=−1alsowehaveF(x)=x+1=t1(t−1)3t4letfindD3(0)forg(t)=(t−1)−3g(t)=g(0)+g′(0)1!t+g(2)(0)2!t2+g(3)(0)3!t3+t44!ξ(t)g(0)=−1,g′(t)=−3(t−1)−4⇒g′(0)=−3g(2)(t)=12(t−1)−5⇒g(2)(0)=−12,g(3)(t)=−60(t−1)−6⇒g(3)(0)=−60⇒g(t)=−1−3t−6t2−10t3+t44!ξ(t)⇒g(t)t4=−1t4−3t3−6t2−10t+14!ξ(t)=−1(x+1)4−3(x+1)3−6(x+1)2−10(x+1)+14!ξ(x+1)⇒b1=−10,b2=−6,b3=−3,b4=−1andF(x)=10x−4x2−1x3−10(x+1)−6(x+1)2−3(x+1)3−1(x+1)4⇒∑n=1∞(−1)nn3(n+1)4=∑n=1∞(−1)nF(n)=10∑n=1∞(−1)nn−4∑n=1∞(−1)nn2−10∑n=1∞(−1)nn+1−6∑n=1∞(−1)n(n+1)2−3∑n=1∞(−1)n(n+1)3−∑n=1∞(−1)nn4
Commented by maxmathsup by imad last updated on 30/May/19
Σ_(n=1) ^∞ (((−1)^n )/n) =−ln(2) Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(3/4)ξ(2) =(3/4) (π^2 /6) =(π^2 /8) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 ) =(π^2 /(24)) −(π^2 /8) =((−2π^2 )/(24)) =−(π^2 /(12)) Σ_(n=1) ^∞ (((−1)^n )/(n+1)) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n) =−Σ_(n=2) ^∞ (((−1)^n )/n) =−{Σ_(n=1) ^∞ (((−1)^n )/n) +1) =−(−ln(2))−1 =ln(2)−1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^2 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^2 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^2 ) =−{ Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +1) =−(−(π^2 /(12)))−1 =(π^2 /(12)) −1 Σ_(n=1) ^∞ (((−1)^n )/((n+1)^3 )) =Σ_(n=2) ^∞ (((−1)^(n−1) )/n^3 ) =−Σ_(n=2) ^∞ (((−1)^n )/n^3 ) for that let calculate δ(x) =Σ_(n=1) ^∞ (((−1)^n )/n^x ) interms of ξ(x) =Σ_(n=1) ^∞ (1/n^x ) (x>1) we have δ(x) =Σ_(n=1) ^∞ (1/(2^x n^x )) − Σ_(n=0) ^∞ (1/((2n+1)^x )) ξ(x) =Σ_(n=1) ^∞ (1/n^x ) =(1/2^x ) ξ(x) +Σ_(n=0) ^∞ (1/((2n+1)^x )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^x )) =(1−2^(−x) )ξ(x) ⇒ δ(x) =2^(−x) ξ(x)−(1−2^(−x) )ξ(x) =(2^(1−x) −1)ξ(x) Σ_(n=1) ^∞ (((−1)^n )/n^3 ) =(2^(−2) −1)ξ(3)=((1/4)−1)ξ(3) =−(3/4)ξ(3) Σ_(n=1) ^∞ (((−1)^n )/n^4 ) =(2^(−3) −1)ξ(4) =((1/8) −1)ξ(4) =−(7/8)ξ(4) so the value of S is determined...
∑n=1∞(−1)nn=−ln(2)∑n=1∞(−1)nn2=14∑n=1∞1n2−∑n=0∞1(2n+1)2∑n=1∞1n2=14∑n=1∞1n2+∑n=0∞1(2n+1)2⇒∑n=0∞1(2n+1)2=34ξ(2)=34π26=π28⇒∑n=1∞(−1)nn2=π224−π28=−2π224=−π212∑n=1∞(−1)nn+1=∑n=2∞(−1)n−1n=−∑n=2∞(−1)nn=−{∑n=1∞(−1)nn+1)=−(−ln(2))−1=ln(2)−1∑n=1∞(−1)n(n+1)2=∑n=2∞(−1)n−1n2=−∑n=2∞(−1)nn2=−{∑n=1∞(−1)nn2+1)=−(−π212)−1=π212−1∑n=1∞(−1)n(n+1)3=∑n=2∞(−1)n−1n3=−∑n=2∞(−1)nn3forthatletcalculateδ(x)=∑n=1∞(−1)nnxintermsofξ(x)=∑n=1∞1nx(x>1)wehaveδ(x)=∑n=1∞12xnx−∑n=0∞1(2n+1)xξ(x)=∑n=1∞1nx=12xξ(x)+∑n=0∞1(2n+1)x⇒∑n=0∞1(2n+1)x=(1−2−x)ξ(x)⇒δ(x)=2−xξ(x)−(1−2−x)ξ(x)=(21−x−1)ξ(x)∑n=1∞(−1)nn3=(2−2−1)ξ(3)=(14−1)ξ(3)=−34ξ(3)∑n=1∞(−1)nn4=(2−3−1)ξ(4)=(18−1)ξ(4)=−78ξ(4)sothevalueofSisdetermined…

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