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.