How Do I Prove That the Limit of \( n!^{n^{-n}} \) Equals 1?

[bomba923]

How do I show that $$\mathop {\lim }\limits_{n \to \infty } \left( {n!} \right)^{\left( {n^{ - n} } \right)} = 1$$ ?

(The n^-n forces the value to decrease faster than n! increases, I believe. But how to work out that?)

 

[estalniath]

1)Firstly we'll let the expression (n!)^(n^-n)=y, then taking ln on both sides give,
lny=ln(n!)/n^n.
2) We'll then find the limit of the expression at the RHS using Le Hopital's rule or by observation that n^n increases faster than ln(n!). =)
3) After finding the limit, L, all we have to do is substitute back the value of y, which is e^L

 

[tacman]

To show that the limit of n!^(n^-n) is equal to 1, we can use the definition of a limit and the properties of exponential and factorial functions.

First, let's rewrite the expression as (n!)^(1/n^n). We can do this by taking the inverse of the exponent, which is equivalent to taking the nth root of both sides.

Now, using the fact that the nth root of a number approaches 1 as n approaches infinity, we can rewrite the expression as (n!)^(1/n^n) = (n!)^(1/n)^(1/n). As n approaches infinity, the first term (n!)^(1/n) also approaches 1, since the factorial function grows much faster than the exponent 1/n.

Therefore, we are left with the limit of 1^(1/n), which is equal to 1. This shows that the limit of n!^(n^-n) is indeed equal to 1.

In other words, the exponential term (n^-n) is decreasing at a much faster rate than the factorial term (n!), and as n approaches infinity, the exponential term becomes negligible compared to the factorial term. This is why the overall limit is equal to 1.

In summary, the limit of n!^(n^-n) = 1 can be shown by using the definition of a limit and the properties of exponential and factorial functions. As n approaches infinity, the exponential term becomes negligible compared to the factorial term, resulting in a limit of 1.