Limits involving factorials $\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$

Question

I am trying to calculate the following limit

$$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$$ where $k$ can be any number between $0$ and $N$.

I thought of the following:

If I take the logarithm of the expression then I get:

$$\lim_{N\to\infty} \left(\log(N!)-\log((N-k)!)-k\log N\right)$$

Using the Stirling formula this can be approximated as:

$$\lim_{N\to\infty} \left(N\log(N)-(N-k)\log(N-k)-k\log N\right)$$

Now there are two cases: If $k$ is $N$, then the second term vanishes and the remaining terms cancel. If $k$ is smaller than $N$, then I can drop the $k$ inside the second logarithm and all the terms cancel. So the limit $$\lim_{N\to\infty} \log\left(\frac{N!}{(N-k)!N^{k}}\right)=0$$ Which means that: $$\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}=1$$ I don't know if this is mathematically rigorous. Would like some help. Thanks

$$\frac{n!}{(n-k)!,n^k}=\frac{n^k+\mathcal O(n^{k-1})}{n^k}\ldots$$ — DonAntonio, Oct 15 '13 at 11:23
It is not clear what $k$ is. Is a fixed integer number? Does it depends on $N$ instead? — ajotatxe, Jun 18 '15 at 13:37
Presumably $k$ is a fixed value. As $N \to \infty$, you will eventually have $N \gt k$, so you only have to worry about that case. — Ross Millikan, Jun 18 '15 at 13:45
See also: Why does $\lim_{n\to\infty} \frac{n!}{(n-k)!n^k}$ equal 1, Finding limit of sequence: $\lim _{n \to \infty} {\frac{n!}{n^k(n-k)!}}=1$, Proof that $\lim\limits_{h \to \infty} \frac{h!}{h^k(h-k)!}=1$ for any $k$, Why $\lim_{n\rightarrow \infty} \frac{n!}{n^{k}(n-k)! } =1$?. — Martin Sleziak, Apr 13 '19 at 07:50

score 12 · Answer 1 · answered Oct 15 '13 at 11:20

12

$$\frac{(n!)}{(n-k)!n^k}=\frac{n(n-1)(n-2)\cdots\{n-(k-1)\}}{n^k}=\prod_{0\le r\le k-1}\left(1-\frac rn\right)$$

answered Oct 15 '13 at 11:20

lab bhattacharjee

274,582

Nice solution lab bhattacharjee.... – juantheron Nov 04 '13 at 02:07
@juantheron, thanks for your feedback. – lab bhattacharjee Nov 05 '13 at 15:46

score 9 · Answer 2 · answered Oct 15 '13 at 11:27

A (non-rigorous) probabilistic interpretation: There are balls of $n$ different colors, of which we choose $k$. The probability that we choose $k$ different colored balls is $$\frac{n(n-1)\dots(n-(k-1))}{n^k} = \frac{n!}{(n-k)!n^k}.$$ It is intuitively clear that as the number of colors increases, this probability goes to $1$.

score 5 · Answer 3 · edited Apr 13 '19 at 08:02

5

I myself just found a better method:

$\frac{N!}{(N-k)!N^k}=\frac{\prod\limits_{i=0}^{k-1} (N-i)}{N^k}$

The top therefore is a polynomial of order k with the highest order having coefficient 1. Therefore all other terms vanish and you are left with 1 as N approaches infinity,

Thanks!

edited Apr 13 '19 at 08:02

Martin Sleziak

53,687

answered Jun 18 '15 at 13:57

eeqesri

91

Good to know that you made it ! Thanks for posting your solution. – Claude Leibovici Jun 18 '15 at 13:58

Limits involving factorials $\lim_{N\to\infty} \frac{N!}{(N-k)!N^{k}}$

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