题目不明确,应为 ∑ (-1)^n [(n+1)!/n^(n-1)] 吧!
ρ = lim|a/a|
= lim(n+2)! n^(n-1)/[(n+1)^n (n+1)!]
= lim(n+2) n^(n-1)/[(n+1)^n ]
= lim(n+2)/(n+1) lim[n/(n+1)]^(n-1)
= 1* lim{[1-1/(n+1)]^[-(n+1)]}^[-(n-1)/(n+1)]
= e^lim -(n-1)/(n+1) = e^lim -(1-1/n)/(1+1/n) = 1/e < 1.
原级数绝对收敛。