Question

幂级数 (x^n)/(n+1) ；求在区间（-1,1）内的和函数S(x)

Answer 1

an=x^n/n+1=x^(n+1)/x(n+1)=(1/x) x^(n+1)/(n+1)

设bn=x^(n+1)/(n+1)

an=(1/x)*bn

S(bn)=∑(x^(n+1)/(n+1)

S(bn)'=∑ x^n= x*(1-x^n)/(1-x)=x/(1-x)

S(bx)=∫xdx/(1-x)=∫-dx+∫dx/(1-x)=-x-ln|1-x|

S(x)=x*S(bx)=-1+(-1/x)ln|1-x|