Simplifying limit with Stirling approximation

[lholmes135]

TL;DR

Using stirling approximation to determine limit at infinity is not giving the correct answer.

I'm trying to determine why

$$ \lim_{N \rightarrow +\infty} ln( \frac {N!} {(N-n)! N^n}) = 0$$

N and n are both positive integers, and n is smaller than N. I want to use Stirling's, which becomes exact as N->inf:

$$ ln(N!) \approx Nln(N)-N $$

And take it term by term:

$$ \lim_{N \rightarrow +\infty} ln(N!) + \lim_{N \rightarrow +\infty} ln((N-n)!) + n \lim_{N \rightarrow +\infty} ln(N))$$
$$ \lim_{N \rightarrow +\infty} [Nln(N)-N] + \lim_{N \rightarrow +\infty} [(N-n)ln(N-n) - (N-n)] + n \lim_{N \rightarrow +\infty} ln(N))$$

I don't really know where to go from here. I tried using the approximation (N-n) = N:

$$ \lim_{N \rightarrow +\infty} [Nln(N)-N] + \lim_{N \rightarrow +\infty} [Nln(N) - N] + n \lim_{N \rightarrow +\infty} ln(N))$$
$$ \lim_{N \rightarrow +\infty} [Nln(N)-N + Nln(N) - N + nln(N))]$$
$$ \lim_{N \rightarrow +\infty} [(2N(ln(N)-1)+nln(N)]$$

But that limit goes to infinity.

 

[Mark44]

Summary:: Using stirling approximation to determine limit at infinity is not giving the correct answer.

I'm trying to determine why

$$ \lim_{N \rightarrow +\infty} ln( \frac {N!} {(N-n)! N^n}) = 0$$

N and n are both positive integers, and n is smaller than N. I want to use Stirling's, which becomes exact as N->inf:

$$ ln(N!) \approx Nln(N)-N $$

And take it term by term:

$$ \lim_{N \rightarrow +\infty} ln(N!) + \lim_{N \rightarrow +\infty} ln((N-n)!) + n \lim_{N \rightarrow +\infty} ln(N))$$

In the equation above, you have two incorrect signs.
$$\ln \frac A {BC} = \ln A - \ln(BC) = = \ln A - (\ln B + \ln C) = \ln A - \ln B - \ln C$$
In your equation the 2nd and 3rd terms should be subtracted, not added.

$$ \lim_{N \rightarrow +\infty} [Nln(N)-N] + \lim_{N \rightarrow +\infty} [(N-n)ln(N-n) - (N-n)] + n \lim_{N \rightarrow +\infty} ln(N))$$

I don't really know where to go from here. I tried using the approximation (N-n) = N:

$$ \lim_{N \rightarrow +\infty} [Nln(N)-N] + \lim_{N \rightarrow +\infty} [Nln(N) - N] + n \lim_{N \rightarrow +\infty} ln(N))$$
$$ \lim_{N \rightarrow +\infty} [Nln(N)-N + Nln(N) - N + nln(N))]$$
$$ \lim_{N \rightarrow +\infty} [(2N(ln(N)-1)+nln(N)]$$

But that limit goes to infinity.

 

[lholmes135]

Thanks Mark, good catch. So fixing the signs and using the same N-n = N approximation we have:

$$ \lim_{N \rightarrow +\infty} {[Nln(N)-N]} - \lim_{N \rightarrow +\infty} {[(N-n)ln(N-n) - (N-n)]} - n \lim_{N \rightarrow +\infty} {ln(N)} $$
$$ \lim_{N \rightarrow +\infty} {[Nln(N)-N]} - \lim_{N \rightarrow +\infty} {[Nln(N) - N]} - n \lim_{N \rightarrow +\infty} {ln(N)} $$
$$ - n \lim_{N \rightarrow +\infty} {ln(N)} $$
$$ -\infty $$

I still can't get 0 out of this.

 

[Stephen Tashi]

N and n are both positive integers, and n is smaller than N. I want to use Stirling's, which becomes exact as N->inf:

Stirlings approximation does not become "exact" as $N \rightarrow \infty$.

i.e $lim_{N \rightarrow \infty} ( N \ln(N) - N - ln(N!) ) \ne 0$

 

[Mark44]

Thanks Mark, good catch. So fixing the signs and using the same N-n = N approximation we have:

$$ \lim_{N \rightarrow +\infty} {[Nln(N)-N]} - \lim_{N \rightarrow +\infty} {[(N-n)ln(N-n) - (N-n)]} - n \lim_{N \rightarrow +\infty} {ln(N)} $$
$$ \lim_{N \rightarrow +\infty} {[Nln(N)-N]} - \lim_{N \rightarrow +\infty} {[Nln(N) - N]} - n \lim_{N \rightarrow +\infty} {ln(N)} $$

The step above isn't valid. In essence, with regard to the 1st and 2nd terms, it is saying that $\infty - \infty = 0$, which isn't necessarily true. As a limit, the indeterminate form $[\infty - \infty]$ can result in a limiting value anywhere between negative and positive infinity.
Expanding the 2nd term, you get $N\ln(N - n) - n\ln(N - n)$
Is it a requirement in this problem to use Stirling's approximation? Evaluating the limit would be simpler just to work with it as it is.

$$ - n \lim_{N \rightarrow +\infty} {ln(N)} $$
$$ -\infty $$

I still can't get 0 out of this.

 

[mathman]

Stirling's approximation is $m!\approx \sqrt{2\pi m}(\frac{m}{e})^m$ Try using it.

 

[Stephen Tashi]

And take it term by term:

Proceeding that way, term by term is not valid.

Regardless of which version of Stirlings approximation you use, it isn't valid to to replace each factorial by it's Stirling approximation because, as indicated previously, if $S(N!)$ a Stirlings approximation for $N!$ the descrepancy between $S(N!)$ and $N!$ approaches $\infty$ as $N \rightarrow\infty$

Stirlings approximation is an asymptotic approximation. So the only valid way to use it is in the form $\lim_{N \rightarrow \infty} \frac{N!}{S(N!)} = 1$ or $\lim_{N \rightarrow \infty} \frac{S(N!}{N!} = 1$.

If you are required to use Stirlings approximation, you should look for ratios in the problem that resemble the above two fractions.

 

[lholmes135]

I suppose I should add some context. This isn't homework. My textbook is deriving a certain formula and I'm trying to follow the derivation. At one step they say something like "and obviously we can use the Stirling formula to show that ..." and show the equation in question. Mathman has posted a better version of the Stirling formula, but the crude version my textbook uses is $$ln(N!) = Nln(N) - N $$

That said, it isn't really a requirement to use the formula. I just want to understand any derivation.

@Stephen Tashi By "exact" I more meant that $$ \lim_{n \rightarrow +\infty} {\frac {Nln(N)-N} {ln(N!)}} = 1 $$
Regardless, I think I see the point you are making that I can't replace it in the expression, and Mark's that I can't replace (N-n) with N.

Also I've been out of school for four years, and this is the first serious math problem I've worked on in that entire time. Hence why I'm doing stupid stuff like mixing up my signs.

 

[lholmes135]

I think I have an explanation without the Stirling approximation.

$$ \lim_{n \rightarrow +\infty} { \frac {N!} {(N-n)!N^n} } $$
$$ \lim_{n \rightarrow +\infty} { \frac {N*(N-1)*...*(N-n+1)} {N*N*...*N} } $$

There are n terms in both the numerator and denominator, so this can be written as

$$ \lim_{n \rightarrow +\infty} { \frac {N} {N} * \frac {N-1} {N} * ... * \frac {N-n+1} {N} } = 1$$
$$ \lim_{n \rightarrow +\infty} { \frac {N!} {(N-n)!N^n} } = 1$$
$$ \lim_{n \rightarrow +\infty} { ln ( \frac {N!} {(N-n)!N^n} )} = 0$$

I appreciate all the help, thank you guys.

 

[Mark44]

Based on the original problem, all of the limits in the previous post should be as $N \to \infty$, not as $n \to \infty$.