Prove $\binom{N}{k}=\frac{N^k}{k!}\left(1+O(\frac{1}{N})\right)$ when $N \rightarrow\infty$

Question

Can someone help prove the following:

$\binom{N}{k}=\frac{N^k}{k!}\left(1+O(\frac{1}{N})\right)$ when $N \rightarrow\infty$.

Thanks in advance!

Answer 1

By definition of $\binom{N}{k}$:

$\binom{N}{k} = \frac{N!}{k!(N-k)!} = \frac{(N-k+1)(N-k+2)\cdots N}{k!}$

by expanding the numerator, we get:

$(N-k+1)(N-k-1)...N = N^k + c_1(k)N^{k-1} + c_2(k)N^{k-2} + \ldots + c_k(k) $

where $c_i(k)$ are polynomial functions, dependent only on $k$. From this we can extract $N^k$ to get:

$N^k\left(1 + \frac{c_1(k)}{N} + \frac{c_2(k)}{N^2} + \ldots + \frac{c_k(k)}{N^k}\right) =N^k\left(1+\mathcal{O}\left(\frac{1}{N}\right)\right)$.

And thus:

$\binom{N}{k} = \frac{N^k}{k!}\left(1+O\left(\frac{1}{N}\right)\right)$