Supremum or upper bound of bivariate function involving logarithms and combinatorial coefficients or the gamma function over a region of the integers

Question

This is a repost from MSE because I got no answers there.

I have been trying to find the supremum of this bivariate function over a specific region. However, the expressions that I get are horrible. I tried Mathematica, but it did not provide a good answer as it gives you something that is clearly not the supremum, as it is smaller than some values I experimented. I hope someone here can help me. I am searching for an upper bound for the function $$\frac{\log\left(\frac{\binom{n}{d}}{2^{d}4}\right)}{d\log(\frac{n}{d})}$$ over the region $\{(n,d)\in\mathbb{N}\mid n>d>0\}.$ I do not need specifically the supremum (although, of course, that would be the best), any (constant) upper bound would also be helpful.

I tried feeding Mathematica with a real extension of this function developing the binomial coefficient using Euler's $\Gamma$ function, which is smooth over the region I am interested in, but Mathematica fails to provide a working upper bound: it gives something that I know it is not an upper bound because I know points in the region having higher value.

Could you help me bounding this function over the region of the integers provided? Does it in fact diverge? If so, how can I see that it diverges?

Answer 1

Note that
\begin{equation*} \frac{\ln\frac{\binom{n}{d}}{2^{d}4}}{d\ln\frac{n}{d}} \le r(n,d):=\frac{\ln\frac{\binom{n}{d}}{2^{d}}}{d\ln\frac{n}{d}}; \end{equation*} here and in what follows, $n$ and $d$ are integers such that $1\le d\le n-1$, as in the OP. So, it is enough to bound $r(n,d)$ from above by a universal constant.

We have \begin{equation*} \binom{n}{d}\le\sqrt{\frac{n}{2\pi d(n-d)}}\,2^{nH((d/n))}, \end{equation*} where $H(p):=-p\log_2 p-(1-p)\log_2(1-p)$. So, letting \begin{equation*} u:=\frac nd, \end{equation*} we get \begin{equation*} d\ge\max(1,\tfrac1{u-1}) \tag{1}\label{1} \end{equation*} and \begin{equation*} r(n,d)\le r_1(d,u)+1+r_3(u), \tag{2}\label{2} \end{equation*} where \begin{equation*} r_1(d,u):=\frac{\ln\sqrt{\frac{n}{2\pi d(n-d)}}}{d\ln\frac{n}{d}} =\frac12\frac{\ln\frac u{u-1}-\ln d}{d\ln u} \end{equation*} and \begin{equation*} r_3(u):=\frac{g(u)-\ln2}{\ln u},\quad g(u):=(u-1)\ln\frac u{u-1}. \end{equation*} Given \eqref{1}, we have the elementary inequalities $r_1(d,u)\le1/2$ and $r_3(u)\le1/2$; see details below.

So, by \eqref{2}, $r(n,d)\le2$. $\quad\Box$

Details on the inequality $r_1(d,u)\le1/2$: Note that $r_1(d,u)$ is decreasing in $d$ such that $r_1(d,u)>0$. So, by \eqref{1}, (i) $r_1(d,u)\le r_1(1,u)=\frac12\frac{\ln u-\ln(u-1)}{\ln u}\le\frac12$ if $u-1\ge1$ and (ii) $r_1(d,u)\le r_1(\tfrac1{u-1},u)=\frac12\,(u-1)\le\frac12$ if $u-1\le1$.

Details on the inequality $r_3(u)\le1/2$: Note that $g''(u)=\frac1{u^2(1-u)}<0$ (for $u>1$), so that $g$ is strictly concave. Also, $g(u)\to1$ as $u\to\infty$. So, $g$ is increasing and $g<1$ . Also, $g(2)=\ln2$ and hence for $u\in(1,2]$ we have $g(u)\le\ln2$, so that $r_3(u)\le0\le1/2$. On the other hand, for $u\ge2$ we have $r_3(u)\le\frac{1-\ln2}{\ln u}\le\frac{1-\ln2}{\ln2}=0.44\ldots<1/2$.