判定级数∑(n=1,∞)(-1)n(n+1)!/n^n-1是否收敛 是绝对收敛还是条件收敛
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题目不明确,应为 ∑<n=1,∞> (-1)^n [(n+1)!/n^(n-1)] 吧!
ρ = lim<n→∞>|a<n+1>/a<n>|
= lim<n→∞>(n+2)! n^(n-1)/[(n+1)^n (n+1)!]
= lim<n→∞>(n+2) n^(n-1)/[(n+1)^n ]
= lim<n→∞>(n+2)/(n+1) lim<n→∞>[n/(n+1)]^(n-1)
= 1* lim<n→∞>{[1-1/(n+1)]^[-(n+1)]}^[-(n-1)/(n+1)]
= e^lim<n→∞> -(n-1)/(n+1) = e^lim<n→∞> -(1-1/n)/(1+1/n) = 1/e < 1.
原级数绝对收敛。
ρ = lim<n→∞>|a<n+1>/a<n>|
= lim<n→∞>(n+2)! n^(n-1)/[(n+1)^n (n+1)!]
= lim<n→∞>(n+2) n^(n-1)/[(n+1)^n ]
= lim<n→∞>(n+2)/(n+1) lim<n→∞>[n/(n+1)]^(n-1)
= 1* lim<n→∞>{[1-1/(n+1)]^[-(n+1)]}^[-(n-1)/(n+1)]
= e^lim<n→∞> -(n-1)/(n+1) = e^lim<n→∞> -(1-1/n)/(1+1/n) = 1/e < 1.
原级数绝对收敛。
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