The target series is S = sum_{n=0}^{infty}n^3((x+1)/2)^nS=∞∑n=0n3(x+12)n with generic term
t=n^3((x+1)/2)^nt=n3(x+12)n
Let us introduce S_b = sum_{n=0}^{infty}((x+1)/2)^nSb=∞∑n=0(x+12)n with generic term t_b=((x+1)/2)^ntb=(x+12)n
S_bSb is convergent for -3 < x < 1−3<x<1. Comparing terms we have
D_3=d^3/(dx^3)t_b=(n(n-1)(n-2))/(2^3)((x+1)/2)^{n-3} = D3=d3dx3tb=n(n−1)(n−2)23(x+12)n−3=
n^3/(2^3)((x+1)/2)^{n-3}-(3n^2)/(2^3)((x+1)/2)^{n-3}+(2n)/(2^3)((x+1)/2)^{n-3}n323(x+12)n−3−3n223(x+12)n−3+2n23(x+12)n−3
then
n^3((x+1)/2)^n = 2^3((x+1)/2)^3d^3/(dx^3)t+3n^2((x+1)/2)^n-2n((x+1)/2)^nn3(x+12)n=23(x+12)3d3dx3t+3n2(x+12)n−2n(x+12)n
Also
D_2=d^2/(dx^2)t_b = (n(n-1))/n((x+1)/2)^{n-2} D2=d2dx2tb=n(n−1)n(x+12)n−2 and analogously
2^2((x+1)/2)^2d^2/(dx^2)t_b = n^2((x+1)/2)^n-n((x+1)/2)^n22(x+12)2d2dx2tb=n2(x+12)n−n(x+12)n
2((x+1)/2)d/(dx)t_b = n((x+1)/2)^n2(x+12)ddxtb=n(x+12)n
2((x+1)/2)D_1= n((x+1)/2)^n2(x+12)D1=n(x+12)n
Putting all together
S_0 = sum_{n=0}^{infty}{2^3 ((x + 1)/2)^3 D_3 + 3 xx 2^2 ((x + 1)/2)^2 D_2 + 2 ((x + 1)/2) D_1}S0=∞∑n=0{23(x+12)3D3+3×22(x+12)2D2+2(x+12)D1}
or
S_o= (2 (1 + x) (13 + 10 x + x^2))/(x-1)^4So=2(1+x)(13+10x+x2)(x−1)4 for -infty < x < 1−∞<x<1
Attached is a figure with the comparison between SS and
S_0 =(2 (1 + x) (13 + 10 x + x^2))/(x-1)^4S0=2(1+x)(13+10x+x2)(x−1)4
Note:
The final expression for S_0S0 is obtained considering that D_k, {k = 1,2,3}Dk,{k=1,2,3} is applied on the equivalence S_b equiv 1/(1-((x+1)/2)Sb≡11−(x+12) for -3 < x < 1−3<x<1
so
D_1 = 2/(x-1)^2D1=2(x−1)2
D_2 = -4/(x-1)^3D2=−4(x−1)3
D_3 = 12/(x-1)^4 D3=12(x−1)4
![enter image source here]()
