= 2∑
= 2[∑
= 2[x/(1-x)]' - 1/(1-x) (-1 < x < 1)
= 2[1(1-x)-x(-1)]/(1-x)^2 - 1/(1-x)
= 2/(1-x)^2 - 1/(1-x) = [2-(1-x)]/(1-x)^2 = (1+x)/(1-x)^2. (-1 < x < 1)
高等数学幂级数和函数
(3) S(x) = ∑(2n+1)x^n = ∑(2n+2-1)x^n
= 2∑(n+1)x^n - ∑x^n
= 2[∑x^(n+1)]' - ∑x^n
= 2[x/(1-x)]' - 1/(1-x) (-1 < x < 1)
= 2[1(1-x)-x(-1)]/(1-x)^2 - 1/(1-x)
= 2/(1-x)^2 - 1/(1-x) = [2-(1-x)]/(1-x)^2 = (1+x)/(1-x)^2. (-1 < x < 1)
= 2∑
= 2[∑
= 2[x/(1-x)]' - 1/(1-x) (-1 < x < 1)
= 2[1(1-x)-x(-1)]/(1-x)^2 - 1/(1-x)
= 2/(1-x)^2 - 1/(1-x) = [2-(1-x)]/(1-x)^2 = (1+x)/(1-x)^2. (-1 < x < 1)
