What is the maximum (absolute) value of the binomial coefficient


$\begin{pmatrix}x \\ k\end{pmatrix} = \frac{1}{k!}x(x-1)(x-2)\dotsb(x-k+1)$


for real $x$ in the interval $0 \leq x \leq k-1$?



Maximum value of the binomial coefficient as a polynomial
It's easy to see that the extremum in $(0,1)$ has the same magnitude as the one in $(k-1,k-2)$ and is more extreme than any of the other extrema. The extremum in $(0,1)$ occurs at $x\_0=(1+o(1))/\ln k$ (by looking at the derivative), but I'll only assume it is $\Theta(1/\ln k)$. Write $x\_0=z/\ln k$ and substitute this value into the function like this:
$$f(x\_0) = \frac{|x\_0(x\_0-1)\cdots(x\_0-k+1)|}{k!} 
 = \frac{z}{k\ln k} \exp\left(\sum\_{j=1}^{k-1}\ln\left(1 - \frac{z}{j\ln k}\right)\right)$$
$$ = \frac{ze^{-z}}{k\ln k}(1+o(1)),$$
where the last step comes from expanding the inside log by Taylor series. The maximum of $ze^{-z}$ occurs for $z=1$, so the asymptotic value of the answer is
$$ \frac{e^{-1}+o(1)}{k\ln k}.$$
We don't even need to prove $x\_0=\Theta(1/\ln k)$ in advance since this expansion shows there is a maximum near $z=1$ and there can only be one maximum in $(0,1)$.


ADDED: Using $\ln(1-y)\le -y$, we obtain a rigorous upper bound 
$$f(x\_0) \lt \frac{e^{-1}}{kH\_{k-1}} \lt \frac{e^{-1}}{k\ln k},$$ where $H\_t=\sum\_{i=1}^t \frac1i$. Experimentally, the first bound is less than 10% high for $k\ge 10$.


