Homework Help: Sequence limit (factorial derivative?)

1. Nov 15, 2012

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?

2. Nov 15, 2012

Ray Vickson

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

3. Nov 15, 2012

Zondrina

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

4. Nov 15, 2012

carlosbgois

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?

5. Nov 15, 2012

Zondrina

Can you prove that $n^n > n!$ think about intervals.

6. Nov 15, 2012

carlosbgois

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

7. Nov 15, 2012

dextercioby

You don't see that for arbitrary n

n times n > 1 times 2 times ... times n ?

8. Nov 15, 2012

carlosbgois

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

9. Nov 15, 2012

Ray Vickson

This is not quite enough: you need $n^n / n!$ to be unbounded, not just > 1.

RGV

10. Nov 15, 2012

dextercioby

Can you prove that, for n>3

$$n^n > \frac{n^n}{n!} > n +1$$ ?

11. Nov 18, 2012

carlosbgois

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

12. Nov 19, 2012

dextercioby

I don't think my method for the second inequality (the 1st is obvious) is induction.