求数列 ((ln n)^n)/n!的敛散性,n为正整数
1个回答

设a[n] = (ln(n))^n/n!,则a[n+1]/a[n] = (ln(n+1))^(n+1)/(ln(n))^n·n!/(n+1)!

= ln(n+1)/(n+1)·(ln(n+1)/ln(n))^n

= ln(n+1)/(n+1)·(1+ln(1+1/n))^n

≤ ln(n+1)/(n+1)·(1+1/n)^n (由不等式ln(1+x) ≤ x)

≤ ln(n+1)/(n+1)·e

于是由lim{n→∞} ln(n+1)/(n+1) = 0,得lim{n→∞} a[n+1]/a[n] = 0.

存在N,使n > N时a[n+1]/a[n] < 1/2.

可得a[n] < a[N]/2^(n-N),0 ≤ lim{n→∞} a[n] ≤ lim{n→∞} a[N]/2^(n-N) = a[N]·2^N·lim{n→∞} 1/2^n = 0.

故a[n] = (ln(n))^n/n!收敛到0.

相关问题