Question

求幂级数的和函数 求和n=1到无穷 （n/（n+1））x^n

Answer 1

收敛区间我就不求了，你自复己算下
收敛区间为(-1，1)
设S(x)=∑
nx^制n/(n+1)
两边同乘x
xS(x)=∑
nx^(n+1)/(n+1)
令F(x)=xS(x)=∑
nx^(n+1)/(n+1)
两边同时求导
F'(x)=∑
nx^n
两边同除x
F'(x)/x=∑
nx^(n-1)
令G(x)=F'(x)/x=∑
nx^(n-1)
两边同时积分zd
∫G(x)dx=∑
x^n=x+x²+x³+……
无穷等比级数
=x/(1-x)
所以G(x)=[x/(1-x)]'=(1-x+x)/(1-x)²=1/(1-x)²
F'(x)=x/(1-x)²=x/(x-1)²=(x-1+1)/(x-1)²=1/(x-1)
+1/(x-1)²
积分得
F(x)=ln(1-x)
-
1/(x-1)
+C
因为F(0)=0S(0)=0
得C=-1
则F(x)=ln(1-x)
+
1/(1-x)
-1=ln(1-x)
+
x/(1-x)
所以S(x)=(1/x)
[ln(1-x)
+
x/(1-x)]