# Thread: Limit of a Sequence

1. ## Limit of a Sequence

Let the sequence an be :

an = n!/n^n , and we know that : Lim n-------> infinity n!/n^n = 0 ,

also we know that n! =
Γ(n+1) ,

Now an =
Γ(x+1)/n^n ,

Now , is it possible to find the limit of this sequence when n⇒ infinity , By using L'Hopital Rule by cosidering :

f(x) =
Γ(x+1)/x^x ,

Lim x-------> infinity f(x) = ??? ( = 0 , by matlab)

Note : I did this limit by matlab and the limit was : 0 ,

I know that the derivative of the gamma function is : Γ(x)*psi(x) ,

How could we do this in L'Hopital Rule and simplify that term to get zero finally ? ,

Best regards

2. ## Re: Limit of a Sequence

$n!\le n^{n-1}$ for all $n\ge 1$

$0\le \dfrac{n!}{n^n} \le \dfrac{1}{n}$

Use the Squeeze Theorem.

3. ## Re: Limit of a Sequence

Thanks for replay , you are right , but this way is already known , i am trying to find the limit by using L'Hopital Rule by converting n! to gamma(n+1) .
Thanks again and best regards

4. ## Re: Limit of a Sequence

$\psi^{(0)}(x) = \dfrac{d}{dx}\left( \log \Gamma(x) \right) = \dfrac{\Gamma'(x)}{\Gamma(x)}$

So, basically, you are finding that $\Gamma'(x) = \Gamma(x) \psi^{(0)}(x) = \Gamma'(x)$. That is not terribly useful. This is not in any way a simplification of the problem. Using polygamma functions gives more of the same.

And taking successive derivatives of the denominator: $\dfrac{d}{dn}(n^n) = n^n(\ln n + 1)$ is going to give you an infinite series at the bottom that is similarly increasing in complexity rather than decreasing.

5. ## Re: Limit of a Sequence

Now, something else you could do that might be a valid way of solving the problem:

$\displaystyle \lim_{n \to \infty} \dfrac{n!}{n^n} = \lim_{n \to \infty} \prod_{k = 1}^n \dfrac{k}{n}$

Now, you are looking to show that infinite product approaches zero. I am not sure if this is any easier.

6. ## Re: Limit of a Sequence

Dear SlipEternal

Thank you soooo much for your replay and for your information , this is what i was looking for .

Best regards