Sequence limit (factorial derivative?)

[carlosbgois]

Homework Statement

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

Homework Equations

$lim_{n->∞}\frac{n^{n}}{n!}$

The Attempt at a Solution

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

$lim_{n->∞}\frac{n^{n}}{n!}$

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

 

[STEMucator]

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

 

[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?

 

[STEMucator]

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

 

[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

 

[dextercioby]

You don't see that for arbitrary n

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

 

[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

 

[Ray Vickson]

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

 

[dextercioby]

Can you prove that, for n>3

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

 

[carlosbgois]

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

 

[dextercioby]

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