I am reading "Concrete Mathematics" book about the convergence proof of binomial theorem, on page 163, it says "...It does. Because $ \binom r k = O(k^{-1-r}) $ by equation (5.83) below...", here $r$ is arbitrary and $k$ is an integer.
For reference, on page 210, equation (5.83) is given as $\frac{1}{z!} = \lim_{n \rightarrow \infty} \binom {n+z} n n^{-z}$.
Can anyone please explain how is $\binom r k = O (k^{-1-r})$ derived from equation (5.83) as above?
Thank you very much.
Write
$$r^{\underline{k}} = \prod_{i=0}^{k-1} (r-i) = (-1)^k \prod_{i=0}^{k-1} (i-r) = (-1)^k (k-1-r)^{\underline{k}}$$
Therefore
$$\binom{r}{k} = \frac{r^{\underline{k}}}{k!} = (-1)^k \binom{k-1-r}{k}$$
Now (5.83) implies
$$\binom{k+z}{k} = \operatorname{O}(k^z) \quad\text{as}\quad k\to\infty$$
Plugging in $z = -1-r$ leads to the stated result.