- #1
lholmes135
- 12
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- TL;DR 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:
$$ \lim_{N \rightarrow +\infty} ln(N!) + \lim_{N \rightarrow +\infty} ln((N-n)!) + n \lim_{N \rightarrow +\infty} ln(N))$$
$$ \lim_{N \rightarrow +\infty} [Nln(N)-N] + \lim_{N \rightarrow +\infty} [(N-n)ln(N-n) - (N-n)] + n \lim_{N \rightarrow +\infty} ln(N))$$
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.
$$ \lim_{N \rightarrow +\infty} ln( \frac {N!} {(N-n)! N^n}) = 0$$
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.