using the fact that ((n+1)/2)^n>n!,
prove that n!<e(n/2)^n
[20 pts]inequality
2011-01-19 5:39 am
回答 (6)
2011-01-19 8:27 am
✔ 最佳答案
Since (1+1/n)^n is increasing ,and lim(n->無限) (1+1/n)^n = ehence, e >= (1+1/n)^n
e >= [(n+1)/n]^n
e >= [(n+1)/2]^n / (n/2)^n
e(n/2)^n >= [(n+1)/2]^n
Given ((n+1)/2)^n>n!,
So that, e(n/2)^n > n!
2011-01-20 1:28 am
myisland的 (1 + 1/n)^n = e 應為 (1 + 1/n)^n <= e
始終只有n趨向無限，等式才成立
Henry1989的 (n/2 + 1/2)^n < (n/2)^n + (1/2)^n 不成立，可以二項式定理推翻
始終只有n趨向無限，等式才成立
Henry1989的 (n/2 + 1/2)^n < (n/2)^n + (1/2)^n 不成立，可以二項式定理推翻
2011-01-19 9:39 pm
((n+1)/2)^n>n!
n!< ((n+1)/2)^n
= (n/2 + 1/2)^n
< (n/2)^n + (1/2)^n
< (n/2)^n + (n/2)^n
< (n/2)^n + e(n/2)^n
n!< ((n+1)/2)^n
= (n/2 + 1/2)^n
< (n/2)^n + (1/2)^n
< (n/2)^n + (n/2)^n
< (n/2)^n + e(n/2)^n
2011-01-19 5:50 am
myisland兄,(n/2)^n (1 + 1/n)^n = e(n/2)^n <--呢步點做的?
2011-01-19 5:50 am
2011-01-19 5:47 am
Prove that n! < e(n/2)^n
n! < [(n+1)/2]^nSince [(n+1)/2]^n= (1/2 + n/2)^n= (n/2)^n (1 + 1/n)^n = e(n/2)^nWe conclude that n! < e(n/2)^n
2011-01-19 22:52:03 補充：
= (n/2)^n (1 + 1/n)^n
< e(n/2)^n (since (1 + 1/n)^n < e for any positive integer n)
n! < [(n+1)/2]^nSince [(n+1)/2]^n= (1/2 + n/2)^n= (n/2)^n (1 + 1/n)^n = e(n/2)^nWe conclude that n! < e(n/2)^n
2011-01-19 22:52:03 補充：
= (n/2)^n (1 + 1/n)^n
< e(n/2)^n (since (1 + 1/n)^n < e for any positive integer n)
收錄日期: 2021-04-22 00:52:18
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