Question

求幂级数∑(∞,n=1)n(n-1)x^n+1的收敛半径,求出和函数

求正确答案 不正确随便摘的不要

Answer 1

级数收敛半径 ρ = lim<n→∞>a<n>/a<n+1>
= lim<n→∞>(n+1)n/[n(n-1)] = 1.
S(x) = ∑<n=1,∞> n(n-1)x^(n+1)
= x^3∑<n=1,∞> n(n-1)x^(n-2) =x^3S1(x),
因 ∫<0,x>du∫<0,u>S1(t)dt
= ∫<0,x>du∫<0,u>[∑<n=1,∞> n(n-1)t^(n-2)]dt
= ∫<0,x>du[∑<n=1,∞> nu^(n-1)]
= ∑<n=1,∞> ‍x^n = x/(1-x)，
得 S1(x) = [x/(1-x)]''= [-1+1/(1-x)]''= [1/(1-x)^2]'= 2/(1-x)^3.
则 S(x) = 2x^3/(1-x)^3 (-1<x<1)