级数[无穷,n=1]n(n+1)x^n求级数的收敛域中,并求和函数.（请给出详细过程）

Question

级数[无穷,n=1] n(n+1)x^n
求级数的收敛域中,并求和函数.（请给出详细过程）

Answer 1

lim(n+1)(n+2)/n(n+1)=1,收敛半径为1
当x=1时,limn(n+1)=无穷大,级数发散
当x=-1时,级数n(n+1)(-1)^n发散
所以收敛域为(-1,1)
设和函数S(x)=∑n(n+1)x^n
则∫S(x)dx=∑n∫(n+1)x^ndx=∑nx^(n+1)=∑(n+2)x^(n+1)-2∑x^(n+1)
=[∑∫(n+2)x^(n+1)dx]'-2x^2/(1-x)
=[∑x^(n+2)]'-2x^2/(1-x)
=[x^3/(1-x)]'-2x^2/(1-x)
=(3x^2-2x^3)/(1-x)^2-2x^2/(1-x)
=x^2/(1-x)^2
s(x)=[x^2/(1-x)^2]'=2x/(1-x)^3,-1