(a^n)/n求和a为常数,n从1至无穷的整数,

Question

(a^n)/n求和 a为常数,n从1至无穷的整数,

Answer 1

设s(x)=∑(x^n)/n 则收敛区间为(-1,1)
s'(x)=∑x^(n-1)=(x^n-1)/(x-1)={[(x-1)+1]^n-1}/(x-1)=(x-1)^(n-1)+...+n
s(x)=∫ (x^n-1)/(x-1)=∫ (x-1)^(n-1)+...+n=...=(x-1)^n/n+...+nx
a=+-1时,是调和级数,发散
|a|>=1时,和发散趋于无穷；|a|<1时
和为s(a)