The Central Eulerian Numbers are given by the formula
$$C(n) = \sum_{j=0}^n(-1)^j(n-j)^{2n-1}\binom{2n}{j}$$
This represent the Eulerian Number $E(2n-1, n)$ where
$$E(n, k) = \sum_{j=0}^k(-1)^j(k-j+1)^n\binom{n+1}{j}$$
It is easy to prove that the sum of Eulerian Numbers at level $n$ is $n!$, i.e.: $\sum_{k=1}^nE(n,k)=n!$
What I want to prove is that the ratio between the Central Eulerian Number at level $2n-1$ and the sum of Eulerian Numbers at level $2n-1$ converges to zero, i.e.:
$$\frac{C(n)}{(2n-1)!}\rightarrow 0$$
The conjecture came from computations at Mathematica:
C[n_] := Sum[(-1)^j*(n - j)^(2 n - 1)*Binomial[2 n, j], {j, 0, n}]
N[C[1]/1 !]
1.
N[C[2]/3 !]
0.666667
N[C[3]/5 !]
0.55
N[C[4]/7 !]
0.479365
N[C[5]/9 !]
0.430418
N[C[10]/19 !]
0.306693
N[C[100]/199 !]
0.0976472
N[C[1000]/1999 !]
0.0308996
N[C[10000]/19999 !]
0.00977198
I expanded the sum and I could the see the last part ($j=n$) converging to zero, but in the beginning of the sum ($j=0$) I have something like
$$\frac{n^{2n-1}}{(2n-1)!}$$
I don't think the factorial can grow as fast as something of the form $n^{2n-1}$, I probably need to use a different technique
Using the asymptotic provided on the OEIS page $$C(n)\sim\frac{2^{2n}n^{2n-1}\sqrt3}{e^{2n}}$$ and comparing it with the asymptotic of $(2n-1)!$ $$(2n-1)!\sim\sqrt{2\pi(2n-1)}\left(\frac{2n-1}e\right)^{2n-1}$$ we get $$\lim_{n\to\infty}\frac{C(n)}{(2n-1)!}$$ $$=\lim_{n\to\infty}\frac{2^{2n}n^{2n-1}\sqrt3}{e^{2n}\sqrt{2\pi(2n-1)}\left(\frac{2n-1}e\right)^{2n-1}}$$ $$=\lim_{n\to\infty}\frac{2^{2n}n^{2n-1}\sqrt3}{e\sqrt{2\pi(2n-1)}(2n-1)^{2n-1}}$$ $$=\lim_{n\to\infty}\left(\frac{2n}{2n-1}\right)^{2n-1}\frac{2\sqrt3}{e\sqrt{2\pi(2n-1)}}$$ $$=\lim_{n\to\infty}\frac{2\sqrt3}{\sqrt{2\pi(2n-1)}}=0$$