求幂级数和函数(n=1到无穷大) (2n-1)/2^n * x^(2n-2) 的和函数 凑不到常用的上面Sg=∑(1到+

设f(x)=(2n-1)/2^n*x^(2n-2),则
g(x)=∫f(x)dx=x^(2n-1)/2^n=(1/x)(x^2/2)^n
Sg=∑(1到+∞）g(x)=(1/x)(x^2/2)/(1-x^2/2)=x/(2-x^2)
Sf=(Sg)'=[x/(2-x^2)]'=(x^2+2)/(2-x^2)^2
又lim[(2n+1)/2^(n+1) * x^(2n)]/[(2n-1)/2^n * x^(2n-2)]=(1/2)x^2
当(1/2)x^2