Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
You could try applying Stirling on the factorials implicit in the binomial coefficient... – J. M. May 2 '11 at 17:07
I don't understand the notations $N^L$ and $C_{N+L}^N$. What do those mean? – Mitch May 2 '11 at 17:17
@Mitch: I am taking $N^L$ as the exponential and $C_{N+L}^N$ as the binomial coefficient of $N+L$ choose $N$ – Ross Millikan May 2 '11 at 17:52
In what regime? If $L$ is fixed and $N$ is allowed to grow then the ratio approaches $L!$. – Qiaochu Yuan May 2 '11 at 17:54
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. – 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.

share|cite|improve this answer

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

share|cite|improve this answer

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

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

share|cite|improve this answer
Thank you, this is a good hint as well. – Leo May 2 '11 at 18:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.