## Sequence limit (factorial derivative?)

1. The problem statement, all variables and given/known data

Find the limit of the sequence given by $S_{n}=\frac{n^{n}}{n!}$

2. Relevant equations

$lim_{n->∞}\frac{n^{n}}{n!}$

3. The attempt at a solution

I know the sequence diverges, but that doesn't mean the limit is also ∞, right?

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Recognitions:
Homework Help
 Quote by carlosbgois 1. The problem statement, all variables and given/known data Find the limit of the sequence given by $S_{n}=\frac{n^{n}}{n!}$ 2. Relevant equations $lim_{n->∞}\frac{n^{n}}{n!}$ 3. The attempt at a solution I know the sequence diverges, but that doesn't mean the limit is also ∞, right?
Either the function f(n) = n^n / n! converges, or else f(n) → +∞ or f(n) → -∞ or else f(n) "oscillates" as n → ∞ in such a way that f(n) does not approach a definite value---not even ± ∞. You need to decide which applies here.

RGV

 Ask yourself, which grows faster, the numerator or the denominator.

## Sequence limit (factorial derivative?)

Thank you both. As the numerator grows faster, and it's a divergence sequence, then the limit is +∞. Now, where may I start to formally prove it?

 Can you prove that $n^n > n!$ think about intervals.
 Yes, I can. For instance, I just evaluated $lim_{n->∞}\frac{3^{n}}{(n+3)!}$ as being 0 by showing that $\forall x \geq 0, (n+3)!>3^{n}$. In a similar way, I may show that $n^{n}>n!$ in the same interval. It just seems to me that this method isn't rigorous enough, you know? As an example, 3>1 is true, but that does't mean that $lim_{x->∞}\frac{x}{3x}=0$ Thank you
 Blog Entries: 9 Recognitions: Homework Help Science Advisor You don't see that for arbitrary n n times n > 1 times 2 times ... times n ?
 Yes, I do, but this seems as an intuitive approach, to me. Isn't it? Just to be sure I got the concepts correctly: Let $s_{n}=\frac{3^{n}}{(n+3)!}$. Then, $s_{1}, s_{2}, ..., s_{n}$ is a sequence, and the partial sum is $S_{x}=s_{1}+s_{2}+...+s_{x}$. That being said, when I say I want to know the limit of the sequence $(lim_{n->∞}\frac{3^{n}}{(n+3)!})$, I'm evaluating the "last" term, $s_{n}$, not the sum to the "last" term, $S_{n}$ right? Many thanks

Recognitions:
Homework Help
 Quote by Zondrina Can you prove that $n^n > n!$ think about intervals.
This is not quite enough: you need $n^n / n!$ to be unbounded, not just > 1.

RGV

 Blog Entries: 9 Recognitions: Homework Help Science Advisor Can you prove that, for n>3 $$n^n > \frac{n^n}{n!} > n +1$$ ?

 Quote by dextercioby Can you prove that, for n>3 $$n^n > \frac{n^n}{n!} > n +1$$ ?
May it be done by induction? It clearly holds for n=3, then I assume it also holds for n=j, and show it's also valid for n=j+1. (Sorry, no paper and pen around right now, I'll try it as soon as I can)

Thanks

 Blog Entries: 9 Recognitions: Homework Help Science Advisor I don't think my method for the second inequality (the 1st is obvious) is induction.