Find the sum of (sigma n=1 to infinity) n(n+1)(x^n) for -1<x<1

Let

f(x) = sum( n=1 to inf, n(n+1) x^(n-1) ).


Integrate twice,

integral(f(x)) = sum( n=1 to inf, (n+1) x^n )

integral( integral(f(x)) ) = sum( n=1 to inf, x^(n+1) ) = x/(1-x).


Differentiate twice,

f(x) = [x/(1-x)]'' = [1/(1-x)^2]' = 2 / (1-x)^3.


Finally, (sigma n=1 to infinity) n(n+1)(x^n) = x*f(x) = 2x / (1-x)^3.