The following series arose in some work related to Hilbert Functions of ideals of points
$$\sum_{k = 0}^m (-1)^k {2m+2 \choose k}[2m(m+1)-k(2m+1)]^{2m-1}.$$
Experimentally, this series is always negative for $m > 1$ and decreases incredibly quickly. We only need that this is never $0$. We have tried rational roots, taking first differences, but have been unable to make much progress. Any help or suggestions are appreciated.
The first term in the asymptotic expansion for the sum, which I'll call $S_0$, is $$S_0 \sim -6 * 2^{-(2m+2)}\sqrt{\frac{3}{m(m+1)}} \binom{2m+2}{m+1}(2m-1)!(2m+1)^{2m-1}.$$ It is explicitly negative and there are no oscillatory terms, so eventually the sum is negative, as conjectured. To complete the proof one would need the second term in the series, show that it is sufficient smaller than the dominant term for a particular $n = n_0$ then check explicitly those values for $n<n_0.$ I'm not going to carry out the analysis for the second term for a lousy 3 points which is what this answer will probably get. I will show how to get the dominant term, however.
For a feel for the accuracy, the ratio of the approximation to the actual is 1.010 for n=300.
Through series manipulations $S_0$ can be written in a more convenient form
$$ S_0 = (2m+1)^{2m-1} \sum_{k=1}^n(-1)^{n+k}\binom{2n}{n+k}\big(k-\frac{n}{2n-1}\big)^{2n-3}, \ \ n=m+1. $$
The analysis will be concerned with the approximation obtained by replacing the $n/(2n-1)$ with 1/2, that is
$$ S := \sum_{k=1}^n(-1)^{n+k}\binom{2n}{n+k}\big(k-\frac{1}{2}\big)^{2n-3}. $$
For a function $f(x)=\sum_{k=1}^\infty f_k x^k,$ $$ \sum_{k=1}^n(-1)^{n+k}\binom{2n}{n+k}f_k = [x^0]f(x)/(x/(x-1)^2)^n $$ where the bracket notation for 'coefficient of' has been used and $[x^0]$ means 'constant term.' For $f_k$ choose $(k-1/2)^{-s}$. Then the sum $f(x)$ can be solved in closed form,
$$ \sum_{k=1}^\infty \frac{x^k}{(k-1/2)^s}=\sqrt{x}\big(2^s\ Li_s(\sqrt{x})-Li_s(x)\big),$$ by use of the definition of the polylogarithm and by bisecting the series. Now for integer $p$ it is known that
$$Li_{-p}(x)=(1-x)^{-(p+1)}\sum_{k=0}^{p-1} A(p,k)x^{p-k} $$ where the $A(p,k)$ are the Eulerian numbers. Thus we have
$$ S=[x^n] \frac{\big((1-\sqrt{x})(1+\sqrt{x})\big)^{2n}}{(1-\sqrt{x})^{2n-2}} 2^{-(2n-3)}\sum_{k=0}^{2n-4}A(2n-3,k)(\sqrt{x})^{2n-3-k} \, - $$ $$ [x^n] (1-x)^2\sum_{k=0}^{2n-4}A(2n-3,k)(\sqrt{x})^{2n-3-k+1/2} . $$
Because within the summation on the second line only half-integral powers of $x$ appear, this line is zero. Now make the substitution $x \to x^2$ and switch summation order to get
$$ S=2^{-(2n-3)}[x^{2n-1}] (1-x)^2(1+x)^{2n}\sum_{k=0}^{2n}A(2n-3,2n-3-k)x^k .$$
Multiply the power series and collect term, getting the expression (with no approximation)
$$S=2^{-(2n-3)} \sum_{k=0}^{2n}A(2n-3,2n-3-k)\Big[ \binom{2n}{k+1}-2\binom{2n}{k+2} + \binom{2n}{k+3} \Big].$$
This expression is advantageous because most of the oscillation in the summand has disappeared. To begin the approximation, note that the binomial and Eulerian coefficients are both strongly peaked and reach their maximum values at the center of the range of their second argument. In particular,
$$\binom{2n}{k} \sim \binom{2n}{n} \exp{\Big(-\frac{(n-k)^2}{n}\Big)} ,\, \, (k\approx n)$$ and
$$A(2n-3,2n-3-k) \sim \sqrt{\frac{3}{\pi(n-1)}}(2n-3)!\exp{\Big(-3\frac{(n-1-k)^2}{n-1}\Big)} .$$
The square brackets 3 equations up is a finite difference approximation to $d^2\binom{2n}{k+1}/\,dk^2.$
Collecting and making a summation index shift we have $$ S \sim \sqrt{\frac{3}{\pi(n-1)}}\,(2n-3)!2^{-(2n-3)}\binom{2n}{n}\,T $$ where $$T=\sum_{k=-n}^{n} \exp{\Big(-3\frac{(k-1)^2}{n-1}\Big)}\frac{d^2}{dk^2} \exp{\Big(-\frac{k^2}{n}\Big)} . $$
Extend the limits on the sum to infinity because we want to use the asymptotic formulation of the theta function. To do so write the sum in terms of a parameter $a,$ which will be evaluated at $a=1$.
$$T\sim -\frac{2}{n}\big( 2\frac{d}{da} + 1\big)\sum_{k=-\infty}^{\infty} \exp{\Big(-3\frac{(k-1)^2}{n-1}\Big)} \exp{\Big(-a\frac{k^2}{n}\Big)} . $$
Complete the square in the exponent to find $$\exp{\big(-a\frac{k^2}{n}-3\frac{(k-1)^2}{n-1}\big)}=\exp{\big(-c(k-d)^2 \,+f\big)}$$ where $$c=\frac{3}{n-1}+\frac{a}{n}, \, d=\frac{3}{c(n-1)},\, f=\frac{-a}{n}d .$$
Now $$\sum_{k=-\infty}^{\infty}\exp{\big(-c(k-d)^2\big)} \sim \sqrt{\frac{\pi}{c}} \big(1+2\exp{(-\pi^2/c)} \cos(2 \pi d) \big)$$ and the second term is exponentially small since $c \to 0$.
Doing the operations as specified for T and then expanding as $n \to \infty,$
$$ T \sim \sqrt{\frac{\pi}{n}} \Big( \frac{-3}{4} + \frac{21}{32n} + ... \Big). $$
There is no point in keeping the second term in the previous equation if the other $O(1/n)$ terms are ignored, but the method shows that it's not difficult to do so for sums like $T$.
Collecting for $S$ we get $$ S \sim -6\, \sqrt{ \frac{3}{n(n-1)}} 2^{-2n}\binom{2n}{n}\, (2n-3)! .$$
The top equation for $S_0$ follows from setting $n=m+1$ and incorporating the leading factor of $(2m+1)^{2m-1}.$
