Proof of Convergence: Limit (n!)/(n^n) = 0

[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

 

[HallsofIvy]

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!

 

[thepatientmental]

To prove that the limit as n --> infinity of (n!)/(n^n) is 0, we can use the definition of convergence of a sequence. According to the definition, a sequence {an} converges to a limit a if for any given small number epsilon, there exists a natural number N such that for all n greater than or equal to N, the absolute value of (an - a) is less than epsilon.

In this case, we want to show that as n approaches infinity, the value of (n!)/(n^n) gets closer and closer to 0. So, let's choose an arbitrary small number epsilon and show that there exists a natural number N such that for all n greater than or equal to N, the absolute value of (n!)/(n^n) is less than epsilon.

First, we can simplify the expression (n!)/(n^n) to (1/n)*(2/n)*...*(n/n). Now, since n is approaching infinity, we know that each term in this product will approach 0. So, for any given epsilon, we can choose N to be a natural number greater than 1/epsilon. This means that for all n greater than or equal to N, each term in the product (1/n)*(2/n)*...*(n/n) will be less than or equal to 1/epsilon. Therefore, the absolute value of (n!)/(n^n) will be less than epsilon, which proves that the limit as n --> infinity of (n!)/(n^n) is 0.

In conclusion, we have shown that for any given small number epsilon, there exists a natural number N such that for all n greater than or equal to N, the absolute value of (n!)/(n^n) is less than epsilon. This satisfies the definition of convergence, thus proving that the limit as n --> infinity of (n!)/(n^n) is 0.