# Sequence limit (factorial derivative?)

## 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?

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Ray Vickson
Homework Helper
Dearly Missed

## 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
Homework Helper
Ask yourself, which grows faster, the numerator or the denominator.

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
Homework Helper
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

dextercioby
Homework Helper
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

Ray Vickson
Homework Helper
Dearly Missed
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
Homework Helper
Can you prove that, for n>3

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

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