Question

求幂级数 ∑(n=1,∝) x^n/[n(n+1)] 的和函数

Answer 1

所求为F(x) x∈[0,1]，这样级数才收敛
那么 xF(x) = ∑(n=1,∝) x^(n+1) / [n(n+1)]
[ xF(x) ] '= ∑(n=1,∝) x^n/n
[ xF(x) ] '' = ∑(n=1,∝) x^(n-1) = 1/(1-x) x∈[0,1)
由二阶导积分得： [ xF(x) ] ' = - ln(1-x) + C1，当x=0时[xF(x)]'等于0所以C1=0
xF(x) = (1-x)ln(1-x) - (1-x) + C2当x=0时xF(x)等于0所以C2=1
所以xF(x) = (1-x)ln(1-x) +x
F(x) =[ (1-x)ln(1-x) + x ] /x ∈(0,1)
当独求出F(0) = 0,F(1) = 1

Answer 2

f(x)=∑x^n/[n(n+1)]
求导：f'(x)=∑ x^(n-1)/(n+1)
F=x^2f'(x)= ∑ x^(n+1)/(n+1)
再求导：F'=∑ x^n=x/(1-x)=1/(1-x)-1
积分：F=-ln(1-x)-x
f'(x)=F/x^2=-ln(1-x)/x^2-1/x
再积分：f(x)=ln(1-x)/x+∫x/(1-x)dx-lnx=ln(1-x)/x-ln(1-x)-x-lnx=[ln(1-x)]/x-ln[x(1-x)]-x

Answer 3

最后一步应该是：
f(x)=[ln(1-x)]/x-ln[(1-x)]