Mathematical induction

P(n) : 2 - 1/n ≥ 1/1! + 1/2! + ... + 1/(n-1)! + 1/n! for n is positive integer
When n = 1, L.H.S. = 2 - 1 = 1
R.H.S. = 1/1! = 1
Hence, L.H.S. ≥ R.H.S.
P(1) is true.
Assume P(k) is true, i.e. 2 - 1/k ≥ 1/1! + 1/2! + ... + 1/(k-1)! + 1/k!, for k is a positive integer.
When n = k+1,
R.H.S. = 1/1! + 1/2! + ... + 1/(k-1)! + 1/k! + 1/(k+1)!
<= 2 - 1/k+ 1/(k+1)!
= 2 - [(k+1)(k-1)! - 1]/(k+1)!
= 2 - [k! + (k+1)! - 1]/(k+1)!
= 1 - (k! - 1)/(k+1)!
= 1 - 1/(k+1) + 1/(k+1)!
<= 2 - 1/(k+1)
P(k+1) is true
By Principle of Mathemtical Induction, P(n) is true for all positive integer n.


2011-03-21 22:58:07 補充：
[Note: 1/(k+1)! <= 1]

2011-03-22 07:08:19 補充：
2 - [(k+1)(k-1)! - 1]/(k+1)!
= 2 - [k(k-1)! + 1*(k-1)! - 1]/(k+1)!
= 2 - [k! + (k-1)! - 1]/(k+1)!

2011-03-22 22:34:05 補充：
I think your method is much clearer than mine.
But any way, we just need to set a sandwich and form the one in between, then we can finally conclude that P(k+1) is correct