Any ideas on finding a good estimate/approximation for $\frac AB$ where $A = N^L$ and $B = {N+L\choose N}$?

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    $\begingroup$ You could try applying Stirling on the factorials implicit in the binomial coefficient... $\endgroup$ – J. M. isn't a mathematician May 2 '11 at 17:07
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    $\begingroup$ I don't understand the notations $N^L$ and $C_{N+L}^N$. What do those mean? $\endgroup$ – Mitch May 2 '11 at 17:17
  • $\begingroup$ @Mitch: I am taking $N^L$ as the exponential and $C_{N+L}^N$ as the binomial coefficient of $N+L$ choose $N$ $\endgroup$ – Ross Millikan May 2 '11 at 17:52
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    $\begingroup$ In what regime? If $L$ is fixed and $N$ is allowed to grow then the ratio approaches $L!$. $\endgroup$ – Qiaochu Yuan May 2 '11 at 17:54
  • $\begingroup$ Wow, just the slightest change in notation (from lower case to upper) made me misunderstand. That's not a problem with the notation, but a problem with my reading ability. $\endgroup$ – Mitch May 2 '11 at 17:56

If you expand $B$ as $\frac{(N+L)!}{N!L!}$ and then use Stirling's approximation on the factorials, you will be very close.


$\log(A/B) = \log(L!) - \sum_{j=1}^L \log(1+j/N)$. You can approximate or bound the sum in various ways, depending on your needs.


I'm assuming that by $C_{N+L}^N$ you mean the binomial coefficient $(N+L)!/N!L!$. (I would denote this by ${N+L \choose N}$.)

If you're thinking of $L$ as a constant then you can write this as

$$ {(N+L)(N+L-1) \cdots (N+1) \over L!}. $$

And you can expand out the numerator; you get

$$ {(N^L + {L(L+1) \over 2} N^{L-1} + \cdots) \over L!} $$

  • $\begingroup$ Thank you, this is a good hint as well. $\endgroup$ – Leo May 2 '11 at 18:59

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