# Proof of convergence

1. Mar 15, 2005

### tonebone10

How can I use the following definition to prove that the limit as n --> infinity of (n!)/(n^n) is 0 :

the sequence {an}-->a if
For all epsilon>0, there exists an N element of natural nos. such that for all n> or = to N, abs value(an-a)<epsilon

2. Mar 15, 2005

### HallsofIvy

Staff Emeritus
That's really the hard way!

Any way, notice that n!= n*(n-1)*(n-2)*...*(3)(2)(1) with n factors while
nn= (n)*(n)*(n)*...*(n)(n)(n) also with n factors. That means that
n!/nn can be written as a product of n factors: (n/n)((n-1)/n)((n-2)/n)...(3/n)(2/n)(1/n). Since the first term is 1 and all others are less than 1,
n!/nn< 1/n. It should be easy to find N such that if n> N, |1/n|< epsilon!