TITLE: Maximum value of the binomial coefficient as a polynomial
QUESTION [7 upvotes]: 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 $0 \leq x \leq k-1$?

REPLY [2 votes]: I'm not sure it adds anything beyond what's in Brendan McKay's answer, but I'll flesh out my comment. The absolute value of the binomial coefficient we are interested in may be written (for $x \in (0,1)$ and $k>1$ ) as
$\frac{x}{k}\prod_{i=1}^{k-1}(1 - \frac{x}{i}).$ Applying the AM-GM inequality to the displayed product, this is at most $\frac{x}{k}(1 - \frac{ xH_{k-1}}{k-1})^{k-1},$ which is in turn at most $$\frac{x e^{-H_{k-1}x}}{k}.$$ As Brendan McKay noted, this is at most $\frac{1}{ekH_{k-1}},$ but perhaps the expression $\frac{x}{k}(1 - \frac{ xH_{k-1}}{k-1})^{k-1}$ will give a better estimate for small $k,$ as that takes maximum value when $x = \frac{k-1}{kH_{k-1}},$ and the maximum value attained there is $\frac{1}{kH_{k-1}}(1-\frac{1}{k})^{k}$ (thanks to Emil Jerabek for a correction here).