求问这题如果写证明过程啊···（详细些），感激不尽啊· 5
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只帮你做第一问，后面自己动动脑子吧。
proof:
first show that lg(n!)<nlgn ,it is easy since lg(n!)=lg(n)+lg(n-1)........<nlogn
second show that there exsit K such that k*lg(n!)>nlgn
let K=2
<= log(n^2 * 1^2)>log(n^2)
log[(n-1)^2 * 2^2]>log(n^2) since n> some N
so on
then theroem proofed.
proof:
first show that lg(n!)<nlgn ,it is easy since lg(n!)=lg(n)+lg(n-1)........<nlogn
second show that there exsit K such that k*lg(n!)>nlgn
let K=2
<= log(n^2 * 1^2)>log(n^2)
log[(n-1)^2 * 2^2]>log(n^2) since n> some N
so on
then theroem proofed.
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