$N^L$ vs. ${N+L\choose N}$

Question

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

Answer 1

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

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

Answer 2

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

Answer 3

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