Question Number 104895 by mathmax by abdo last updated on 24/Jul/20

1) decompose the fraction  F(x) =(1/(x^3 (x+1)^4 ))  2) find the sumA =  Σ_(n=1) ^∞  (1/(n^3 (n+1)^4 ))  and B =Σ_(n=1) ^(∞ )  (((−1)^n )/(n^3 (n+1)^4 ))  3) what is the value of  Σ_(n=0) ^∞   (1/((n+1)^4 (2n+1)^3 )) ?

Answered by Ar Brandon last updated on 25/Jul/20

1\ F(x)=(1/(x^3 (x+1)^4 ))=(a/(x+1))+(b/((x+1)^2 ))+(c/((x+1)^3 ))+(d/((x+1)^4 ))+((ex^2 +fx+g)/x^3 )  F(x)=((ax^3 (x+1)^3 +bx^3 (x+1)^2 +cx^3 (x+1)+dx^3 +(ex^2 +fx+g)(x+1)^4 )/(x^3 (x+1)^4 ))  x→−1⇒d=−1, x→0⇒g=1, a+e=_x^6  0, 3a+b+4e+f=_x^5  0,   3a+2b+c+6e+4f+g=_x^4  0, a+b+c+d+4e+6f+4g=_x^3  0,  e+4f+6g=_x^2  0, f+4g=_x 0⇒f=−4, e=10, a=−10, b=−6, c=−3  ⇒F(x)=((−10)/(x+1))+((−6)/((x+1)^2 ))+((−3)/((x+1)^3 ))+((−1)/((x+1)^4 ))+((10x^2 −4x+1)/x^3 )

Answered by mathmax by abdo last updated on 25/Jul/20

1) F(x) =(1/(x^3 (x+1)^4 )) ⇒F(x) =Σ_(i=1) ^3  (a_i /x^i ) +Σ_(i=1) ^4  (b_i /((x+1)^i ))  to find a_(i )  we determine D_2 (0) for f(x) =(1/((x+1)^4 )) =(x+1)^(−4)   f(x) =f(0) +(x/(1!))f^′ (0) +(x^2 /(2!))f^((2)) (0) +(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 /6)ξ(x) ⇒  ((f(x))/x^3 ) =(1/x^3 )−(4/x^2 ) +((10)/x) +(1/6)ξ(x) ⇒ a_1 =10 ,a_2 =−4 , a_3 =1  to find b_i  we determine D_3 (−1) for g(x) =(1/x^3 ) =x^(−3)  ⇒  g(x) =g(−1) +(x+1)g^′ (−1) +(((x+1)^2 )/2)g^((2)) (−1) +(((x+1)^3 )/6)g^((3)) (−1) +(((x+1)^4 )/(4!))ξ(x)  g(−1) =−1  , g^′ (x) =−3x^(−4)  ⇒g^′ (−1) =−3  g^((2)) (x) =12 x^(−5)  ⇒g^((2)) (−1) =−12  ,  g^((3)) (x) =−60 x^(−6)  ⇒g^((3)) (−1) =−60  ⇒g(x) =−1 −3(x+1)−6(x+1)^2  −10(x+1)^3  +(((x+1)^3 )/(3!))ξ(x) ⇒  ((g(x))/((x+1)^4 )) =−(1/((x+1)^4 ))−(3/((x+1)^3 ))−(6/((x+1)^2 ))−((10)/(x+1)) +(1/(4!))ξ(x) ⇒  b_1 =−10 ,b_2 =−6  , b_3 =−3 ,  b_4 =−1 ⇒  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 ))

Commented bymathmax by abdo last updated on 25/Jul/20

2) S =Σ_(n=1) ^∞  (1/(n^3 (n+1)^4 )) ⇒S=lim_(n→+∞) Σ_(k=1) ^n  (1/(k^3 (k+1)^4 ))  we have Σ_(k=1) ^n  (1/(k^3 (k+1)^4 )) =10 Σ_(k=1) ^n  (1/k)−4Σ_(k=1) ^n  (1/k^2 ) +Σ_(k=1) ^n  (1/k^3 )−10 Σ_(k=1) ^n  (1/(k+1))  −6 Σ_(k=1) ^n  (1/((k+1)^2 ))−3 Σ_(k=1) ^n  (1/((k+1)^3 ))−Σ_(k=1) ^n  (1/((k+1)^4 ))  10Σ_(k=1) ^n ((1/k)−(1/(k+1))) =10(1−(1/2)+(1/2)−(1/3)+....+(1/n)−(1/(n+1))) =10(1−(1/(n+1)))→10  Σ_(k=1) ^n  (1/k^2 )→(π^2 /6)  and Σ_(k=1) ^n  (1/k^3 ) →ξ(3)  Σ_(k=1) ^n  (1/((k+1)^2 )) =Σ_(k=2) ^(n+1)  (1/k^2 )→(π^2 /6)−1 and Σ_(k=1) ^n  (1/((k+1)^3 )) =Σ_(k=2) ^(n+1)  (1/k^3 )→ξ(3)−1  Σ_(k=1) ^n  (1/((k+1)^4 )) =Σ_(k=2) ^(n+1)  (1/k^4 ) →ξ(4)−1 ⇒  S =10−4×(π^2 /6) +ξ(3)−6((π^2 /6)−1)−3(ξ(3)−1)−(ξ(4)−1) ⇒  S =10−((2π^2 )/3) +ξ(3)−π^2  +6−3ξ(3)+3−ξ(4)+1  =20−((5π^2 )/3) −2ξ(3)−ξ(4)