I am wondering how to estimate the following summation.
For $p \ge 0$,
$$ \sum_{i=0}^n \binom{2n}{i} (-1)^i (n-i)^p. $$
When $p$ is a fixed integer this seems easy to do. But what if $p$ is a general real number? I performed some numerical simulation and it seems this summation equals zero iff p is even. Can we give some lower bound to this summation, in general?
Update: the lower bound of the summation should be i=0, sorry for the confusion.
On
Working with the question asked about $p$ an even integer we evaluate
$$S_{n,p} = \sum_{q=0}^{n} {2n\choose q} (-1)^q (n-q)^p.$$
Note that
$$\sum_{q=n}^{2n} {2n\choose q} (-1)^q (n-q)^p = \sum_{q=0}^{n} {2n\choose n+q} (-1)^{n+q} (-q)^p \\ = \sum_{q=0}^{n} {2n\choose 2n-q} (-1)^q (q-n)^p.$$
With $p$ even this is
$$\sum_{q=0}^{n} {2n\choose q} (-1)^q (n-q)^p.$$
Since the duplicate term for $q=n$ from the lower and upper sum is zero we thus have
$$2S_{n,p} = \sum_{q=0}^{2n} {2n\choose q} (-1)^q (n-q)^p \\ = p! [z^p] \sum_{q=0}^{2n} {2n\choose q} (-1)^q \exp((n-q)z) \\ = p! [z^p] \exp(nz) \sum_{q=0}^{2n} {2n\choose q} (-1)^q \exp(-qz) \\ = p! [z^p] \exp(nz) (1-\exp(-z))^{2n}.$$
Since $1-\exp(-z) = z + \cdots$ and hence $(1-\exp(z))^{2n} = z^{2n} +\cdots$ this is zero when $p\lt 2n.$ In other words, $S_{n,p} = 0$ when $p$ is even and $2n\gt p.$ We also find that for $p=2n$ the value is
$$\frac{1}{2} (2n)! [z^{2n}] \exp(nz) (1-\exp(-z))^{2n} = \frac{1}{2} (2n)! [z^0] \exp(nz) = \frac{1}{2} (2n)!.$$
Cannot post this as comment: a few graphs for $n=20$ with horizontal axis representing values of $p$.
Here is the part for $8<p<10$:
...and for $10<p<12$: