f(x) = ∑ x^n/（n+1）xf(x) = ∑ [x^(n+1)]/（n+1）[xf(x)]' = ∑ x^n所以[xf(x)]'的和函数很好求,就是等比级数,所以[xf(x)]' = 1/(1-x)所以xf(x) = ∫ 1/(1-x)dx = -ln(1-x) f(x)=-[ln(1-x)]/x,最后协商收敛于x属...