Does the Limit of aCn/na Approach 1/a! as n Approaches Infinity?

[nhmllr]

Here's the weird limit:

lim n\rightarrow \infty _aC_n/n^a = 1/(a!)

Don't ask how I thought this up, but let me explain my reasoning.
Let's say that a=3. Then _aC_n = (n)(n-1)(n-2)
And because n is approaching infinity, could it be (n)(n)(n) = n³?

I wouldn't normally think of things like this but if it's true then it helps me out.

The only thing is that (n)(n-1)(n-2)= n³ +something*n²+something*n

And when I graphed it it looked like it approaches zero.
I've never learned limits formally. Thanks.

EDIT: _aC_n is a Choose n, or n!/(a![n-a]!)
EDIT: Wolfram alpha is amazing http://www.wolframalpha.com/input/?i=lim+(x!/(3!(x-3)!x^3))+as+x->+infinity
Sorry for not trying this before posting. I'd still be interested in other replies as to why this is true, though

 

[HallsofIvy]

Here's the weird limit:

lim n\rightarrow \infty _aC_n/n^a = 1/(a!)

Don't ask how I thought this up, but let me explain my reasoning.
Let's say that a=3. Then _aC_n = (n)(n-1)(n-2)
And because n is approaching infinity, could it be (n)(n)(n) = n³?

No, because n is not fixed, it is, as you said, approaching infinity.
In any case, _aC_n= n(n-1)(n-3)/3!, not just n(n-1)(n-2). That goes to infinity as n goes to infinity.

I wouldn't normally think of things like this but if it's true then it helps me out.

The only thing is that (n)(n-1)(n-2)= n³ +something*n²+something*n

And when I graphed it it looked like it approaches zero.
I've never learned limits formally. Thanks.

EDIT: _aC_n is a Choose n, or n!/(a![n-a]!)
EDIT: Wolfram alpha is amazing http://www.wolframalpha.com/input/?i=lim+(x!/(3!(x-3)!x^3))+as+x->+infinity
Sorry for not trying this before posting. I'd still be interested in other replies as to why this is true, though

Now, this is completely different- you have an n^3 in the denominator.
\frac{_nC_3}{n^3}= \frac{n!}{3!(n-3)!}\frac{1}{n^3}= \frac{n(n-1)(n-2)}{6n^3}
Since both numerator and denominator are of third power in n, that limit is 1/6.

 

[aesir]

I have not tried yet, but a stirling's approximation of {}_aC_n would give the correct limit
n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n

 

[identity1]

No, you're limit is correct. It's also not that weird. Consider the expansion of {n\choose a}=\frac{n(n-1)\ldots (n-a+1)}{a!}. Then, consider that \frac{n(n-1)\ldots (n-a+1)}{n^a}=1\cdot (1-\frac{1}{n})\cdot\ldots\cdot (1-\frac{a-1}{n}). Each of those factors goes to 1 as n\rightarrow\infty, thus the whole product goes to 1 as well. The expression {n\choose a}n^{-a} is the same thing, but divided by a!.