Question

求解答...无穷级数的和

Answer 1

考虑级数f(x)=∑(0,+∞)(-1)^nx^(3n+1)/(3n+1)
求导得：f'(x)=∑(0,+∞)(-1)^nx^(3n)=1/(1+x^3) （|x|<1
积分得：f(x)=∫dx/(1+x^3)
由于1/(1+x^3)=1/(1+x)(1-x+x^2)=(1/3)/(1+x)-(1/6)(2x-1)/(1-x+x^2)+(1/2)/((x-1/2)^2+3/4))
所以：f(x)=(1/3)ln|x+1|-(1/6)ln(1-x+x^2) + (1/√3)arctan((2x-1)/√3)+C
因为级数在x=1收敛，故上式成立范围 -1<x<=1
由于f(0)=0,C=(1/√3)arctan(1/√3)
故∑(0,+∞)(-1)^n/(3n+1)=f(1)= (1/3)ln2+ (2/√3)arctan(1/√3)
=(1/3)ln2+ (2/√3)(π/6)
=(1/3)(ln2+π/√3)