Is there a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$

Question

In a proof of the Larman-Rogers conjecture (there is $\gamma>1$ such that $\chi(\mathbb{R}^{d})>\gamma^d) $ they used that there is a $\gamma>1$ such that $\frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+\cdots+\binom{4p}{p-1}}>\gamma^{p}$. I do not see why this is true. I tried manipulating both parts of the fraction to get the desired result. I also tried to use Stirling's approximation but without success.

EDIT: using $(\frac{n}{k})^k$ $\geq$ $\binom{n}{k}$ $\geq$ $(\frac{en}{k})^k$ seems to help a lot but it is still not getting quite right.

Answer 1

By induction on $n$, using Pascal's rule: $$ \binom{n}{r} = \binom{n - 1}{r} + \binom{n - 1}{r - 1} \quad (n \geqslant r \geqslant 1), $$ we get, for $n > r \geqslant 0$: \begin{equation} \tag{$*$}\label{eq:id} \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} = \binom{n - 1}{r} + 2\binom{n - 2}{r - 1} + \cdots + 2^{r - 1}\binom{n - r}{1} + 2^r. \end{equation} Suppose: $$ \frac{r}{n - 1} \leqslant \frac{1}{4}. $$ Then: $$ \frac{r - k}{n - k - 1} \leqslant \frac{1}{4} \quad (k = 0, 1, \ldots, r), $$ whence: \begin{gather*} \binom{n - k - 1}{r - k} = \left(\frac{r}{n - 1}\right)\left(\frac{r - 1}{n - 2}\right)\cdots\left(\frac{r - k + 1}{n - k}\right)\binom{n - 1}{r} \\ \leqslant 4^{-k}\binom{n - 1}{r} \quad (k = 0, 1, \ldots, r). \end{gather*} From \eqref{eq:id}, therefore: \begin{gather*} \binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{r} \leqslant \left(1 + 2^{-1} + \cdots + 2^{-r+1} + 2^{-r}\right)\binom{n - 1}{r} \\ < 2\binom{n - 1}{r} \quad \left(0 \leqslant r \leqslant \left\lfloor\frac{n - 1}{4}\right\rfloor\right). \end{gather*} In particular: $$\binom{4p}{0} + \binom{4p}{1} + \cdots + \binom{4p}{p - 1} < 2\binom{4p - 1}{p - 1} \quad (p \geqslant 1). $$ Therefore: \begin{gather*} \frac{\binom{4p}{2p-1}}{\binom{4p}{0}+\binom{4p}{1}+…+\binom{4p}{p-1}} > \frac{\binom{4p}{2p - 1}}{2\binom{4p - 1}{p - 1}} = \frac{2p(p - 1)!(3p)!}{(2p - 1)!(2p + 1)!} \\ = 2\left(\frac{2p + 2}{p + 1}\right) \left(\frac{2p + 3}{p + 2}\right)\cdots \left(\frac{3p}{2p - 1}\right) \\ \geqslant 2\left(\frac{3}{2}\right)^{p - 1} \geqslant \left(\frac{3}{2}\right)^p\quad (p \geqslant 1). \end{gather*}

Answer 2

Probaly too complex and almost non rigorous.

$$\sum_{k=0}^{p-1} \binom{4 p}{k}=16^p-\binom{4 p}{p} \, _2F_1(1,-3 p;p+1;-1)$$ where appears the Gaussian or ordinary hypergeometric function.

So, let us consider $$y=\frac{\binom{4 p}{2 p-1}}{16^p-\binom{4 p}{p} \, _2F_1(1,-3 p;p+1;-1)}$$ Computing it, it seems that $\log(y)$ is almost linear with respect to $p$. I generated values for $10 \leq p \leq 10000$ (stepsize equal to $10$) and a quick and dirty linear regression gave for $\log(y)=a+b\,p$ $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & 0.551608 & 0.000208 & \{0.551199,0.552017\} \\ b & 0.523248 & \approx 0 & \{0.523248,0.523248\} \\ \end{array}$$ So, it seems that a lower bound of $\gamma$ is around $1.6875$.

Edit

You have received two very elegant solutions from Calum Gilhooley and from Sangchul Lee.

Considering the first answer where appears $$A=\frac{2\,p\,(p - 1)!\,(3p)!}{(2p - 1)!\,(2p + 1)!}$$ using Stirling approximation, we have $$\log(A)=\frac{\log (3)}{2}+ \log \left(\frac{27}{16}\right)p-\frac{17}{36 p}+O\left(\frac{1}{p^2}\right)$$ and $\frac{\log (3)}{2}\approx 0.549306$ and $\log \left(\frac{27}{16}\right)\approx 0.523248$ which I was unable to identify.

Considering the second answer where appears $$\frac{1}{p}\log\left( \frac{\binom{4p}{2p-1}}{p \binom{4p}{p-1}} \right) \leq \alpha \leq \frac{1}{p}\log\left( \frac{\binom{4p}{2p-1}}{\binom{4p}{p-1}} \right)$$ and doing the same $$\log \left(\frac{27}{16}\right)+\frac{\log \left(\frac{3 \sqrt{3}}{2}\right)-\log(p)}{p}+O\left(\frac{1}{p^2}\right) \leq \log(\alpha) \leq \log \left(\frac{27}{16}\right)+\frac{\log \left(\frac{3 \sqrt{3}}{2}\right)}{p}+O\left(\frac{1}{p^2}\right) $$

Answer 3

1. Here is a modification of @Calum Gilhooley's solution. Notice that $k \mapsto \binom{4p}{k}$ is strictly increasing for $k \in [0, 2p]$. So

   $$ \frac{\binom{4p}{2p-1}}{\sum_{k=0}^{p-1}\binom{4p}{k}} > \frac{\binom{4p}{2p-1}}{p\binom{4p}{p-1}} = \frac{1}{p}\frac{(p-1)!(3p+1)!}{(2p-1)!(2p+1)!} = \frac{1}{p} \prod_{k=0}^{p-1} \frac{2p+2+k}{p+k} $$

   Since $k \mapsto \frac{2p+2+k}{p+k}$ is decreasing in $k$,

   $$ \frac{\binom{4p}{2p-1}}{\sum_{k=0}^{p-1}\binom{4p}{k}} \geq \frac{1}{p} \left( \frac{3p+1}{2p-1} \right)^p \geq \frac{1}{p}\left(\frac{3}{2}\right)^p. $$

   Now it is easy to find $\gamma > 1$ such that $\frac{1}{p}\left(\frac{3}{2}\right)^p \geq \gamma^p$.

2. For the best choice of $\gamma$, numerical evidence suggests that

   $$ \alpha(p) := \frac{1}{p}\log\left( \frac{\binom{4p}{2p-1}}{\sum_{k=0}^{p-1} \binom{4p}{k}} \right) $$

   is strictly decreasing in $p$. So the best possible choice will be $\inf_{p > 0} \alpha(p) = \lim_{p\to\infty} \alpha(p)$. From

   $$ \frac{1}{p}\log\left( \frac{\binom{4p}{2p-1}}{p \binom{4p}{p-1}} \right) \leq \alpha(p) \leq \frac{1}{p}\log\left( \frac{\binom{4p}{2p-1}}{\binom{4p}{p-1}} \right), $$

   it is easy to show that $\lim_{p\to\infty}\alpha(p) = \log \left(\frac{27}{16}\right)$, hence accepting the monotonicity of $\alpha$, we have

   $$ \frac{\binom{4p}{2p-1}}{\sum_{k=0}^{p-1} \binom{4p}{k}} > \left(\frac{27}{16}\right)^p $$

   and no $\gamma > \frac{27}{16}$ will satisfy the inequality for all $p > 0$.