Krdytkyu

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.

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} + 2binom{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)cdotsleft(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} \
< 2binom{n - 1}{r} quad left(0 leqslant r leqslant leftlfloorfrac{n - 1}{4}rightrfloorright).
end{gather*}
In particular:
$$binom{4p}{0} + binom{4p}{1} + cdots + binom{4p}{p - 1} < 2binom{4p - 1}{p - 1}
quad (p geqslant 1).
$$
Therefore:
begin{gather*}
frac{binom{4p}{2p-1}}{binom{4p}{0}+binom{4p}{1}+…+binom{4p}{p-1}}
> frac{binom{4p}{2p - 1}}{2binom{4p - 1}{p - 1}}
= frac{2p(p - 1)!(3p)!}{(2p - 1)!(2p + 1)!} \
= 2left(frac{2p + 2}{p + 1}right)
left(frac{2p + 3}{p + 2}right)cdots
left(frac{3p}{2p - 1}right) \
geqslant 2left(frac{3}{2}right)^{p - 1}
geqslant left(frac{3}{2}right)^pquad (p geqslant 1).
end{gather*}

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

   
   
   

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

   
   
   

   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

   
   
   

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

   
   
   

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

   
   
   

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

   
   
   

   it is easy to show that $lim_{ptoinfty}alpha(p) = log left(frac{27}{16}right)$, 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 $p > 0$.

   
   

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
$$log(A)=frac{log (3)}{2}+ log left(frac{27}{16}right)p-frac{17}{36 p}+Oleft(frac{1}{p^2}right)$$ 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
$$frac{1}{p}logleft( frac{binom{4p}{2p-1}}{p binom{4p}{p-1}} right) leq alpha leq frac{1}{p}logleft( frac{binom{4p}{2p-1}}{binom{4p}{p-1}} right)$$ and doing the same
$$log left(frac{27}{16}right)+frac{log left(frac{3
sqrt{3}}{2}right)-log(p)}{p}+Oleft(frac{1}{p^2}right) leq log(alpha) leq log left(frac{27}{16}right)+frac{log left(frac{3
sqrt{3}}{2}right)}{p}+Oleft(frac{1}{p^2}right) $$