math 問題 ，有關exponmential series

The answer is actually very easy.
There is no need to use M.I. or else. Just Inquality is enough.

n! = n x ( n - 1 ) x ( n - 2 ) ... x 2 x 1
= n x ( n - 1 ) x ( n - 2 ) ... x 2 [ a total of ( n - 1 ) terms ]
> 2 x 2 x 2 .... x 2 [ also a total of ( n - 1 ) terms ]
= 2^( n - 1 )

Taking their reciprocals, we have 1/n! < 1/2^( n - 1 )

That is finished!

Approach 1 用MI:
n=3:
LHS = 2^(3-1) = 4 < 6 = 3! = RHS
=> n=3 ture

Assume n=k true, i.e., 2^(k-1)<n!

For n=k+1,

LHS = 2^[(k+1)-1] = 2( 2^(k-1) ) < 2 (n!) < (n+1)(n!) < (n+1)!
=>n=k+1正確 => By principle of MI, true for all positive integer >2.

Approach 2:
由於
2 = 2
2 < 3
2 < 4
...
2 < n
將以上咁多條乘埋，得
2*2*2,,,*2 < 2*3*4...*n
整理得 2^(n-1) < n!, 得證。

雖然我的證明並不嚴謹，但也可參考。

已知2^(2-1)=2!

因為如在左式*2, 會使左式變為2^(3-1)

　　如在右式*3, 會使左式變為3!

又因為左式和右式均為正數，所以n=3時此證成立


設n=k, 而k>2, 2^(k-1)<k!
設n=k+1,
L.H.S.
=2^[(k+1)-1]
=2^(k-1)*2

R.H.S.
=(k+1)!
=k!*(k+1)

因為 2^(k-1) 和 k! 均為正數, 又因(k+1)>2,
所以L.H.S.<R.H.S.

所以所有大於2的正整數n, 2^(n-1) < n!