求幂级数的和函数。如图

Question

如题所述

Answer 1

级数的导数等于等比级数x^2n的和，等于1/(1-x^2),所以级数等于1/(1-x^2)的从0到x（|x|<1）的积分，即等于ln((1+x)/(1-x))/2

Answer 2

S(x) = ∑<n=1,∞>x^(2n+1)/(2n+1)
S'(x) = ∑<n=1,∞>x^2n = x^2/(1-x) (-1<x<1)
S(x) = ∫<0, x> S'(t)dt - S(0)
= ∫<0, x> t^2/(1-t)dt - 0 = ∫<0, x> t^2/(1-t)dt
= [-t^2/2-t-ln(1-t)]<0, x> = -x^2/2-x-ln(1-x) (-1<x<1)追问

那不是2n-1吗？

追答

S(x) = ∑x^(2n-1)/(2n-1)
S'(x) = ∑x^(2n-2) = 1/(1-x^2) (-1 S'(t)dt - S(0)
= ∫ 1/(1-t^2)dt - 0 = (1/2)∫ [1/(1-t)+1/(1+t)]dt
= (1/2)[ln(1+t)/(1-t)] = (1/2)ln[(1+x)/(1-x)] (-1<x<1)

追问

谢谢你！