I don't have the time to slice through it right now, but this is solved by this.

Your link doesn't work grobber.

That's strange because it works perfectly fine for me. Anyway, it points to the topic named "Sum", with source "I forgot...", posted by Bruno Holanda in the Proposed and Own Problems subsection of the Algebra section.

The link doesn't work for me either, but here's a solution (sorry if it's the same as the one on Grobber's link, for those of you who can view it): $\frac {1}{2n+2}*( (n+k)! (n-k+1)! + (n+k+1)! (n-k)! ) = \frac {1}{2n+2} (n+k)! (n-k)! ((n - k +1) + (n+k+1)) = (n-k)! (n+k)!$. Therefore, $\sum_{k=0}^n (-1)^k (n-k)!(n+k)! = \frac {1}{2n+2}\sum_{k=0}^n (-1)^k ((n+k)! (n-k+1)! + (n+k+1)! (n-k)!) = \frac {1}{2n+2}(n! (n+1)! + (-1)^n (2n+1)!)$, a valid closed form expression.

Yes it is the same I gave for the other problem (and the link works for me ) Pierre.

Let's consider $S = \sum_{k=0}^{2n} (-1)^{k}\frac{k!(2n-k)!}{(2n + 1)!}$. We recognize $\frac{k!(2n-k)!}{(2n + 1)!}$ as $\frac{\Gamma(k+1) \Gamma(2n - k + 1)}{\Gamma(2n + 2)} = \int_{0}^1 x^k (1-x)^{2n-k} \, dx$, so \begin{align*} S &= \sum_{k=0}^{2n}\left( (-1)^k \int_{0}^1 x^k (1-x)^{2n-k} \, dx\right) \\ &= \int_{0}^1\left( \sum_{k=0}^{2n} (-x)^k (1-x)^{2n-k}\right) \, dx \\ &= \int_{0}^1 (1 -x)^{2n} \left(\sum_{k=0}^{2n} \left(\frac{-x}{1-x}\right)^{k} \right)\, dx \\ &= \int_{0}^1 (1-x)^{2n} \frac{\left(\frac{-x}{1-x}\right)^{2n + 1} - 1} {\frac{-x}{1-x} - 1} \, dx \\ &= \int_{0}^1 ((1-x)^{2n + 1} - (-x)^{2n + 1}) \, dx \\ &= \frac{1}{n+1} \end{align*} Now, \begin{align*} \frac{1}{n+1} &= S \\ &= \sum_{k=0}^{2n} (-1)^{k}\frac{k!(2n-k)!}{(2n + 1)!} \\ &= \sum_{k=0}^n (-1)^{k}\frac{k!(2n-k)!}{(2n + 1)!} + \sum_{k=n}^{2n} (-1)^{k}\frac{k!(2n-k)!}{(2n + 1)!} - (-1)^n \frac{n!(2n-n)!}{(2n+1)!} \\ &=\frac{(-1)^n}{(2n+1)!} \left(2\sum_{k=0}^n (-1)^k (n+k)! (n-k)! - (n!)^2\right), \end{align*} (by shifting the indices: $k \mapsto n - k$ in the first sum and $k \mapsto k - n$ in the second) Therefore, $\sum_{k=0}^n (-1)^k (n+k)! (n-k)! = \frac{(2n+1)!(-1)^n}{2(n+1)} + \frac{(n!)^2}{2}$.

Let $ B(x, y) = \frac{x!y!}{(x + y + 1)!} = \int_0^{1}t^x(1 - t)^ydt $ be the Beta function (or an Eulerian integral of the first kind). Lemma: $ \sum_{m = 0}^{n}\frac{(-1)^m}{n + m}\binom{n}{m} = B(n - 1, n) $ Proof: Note that $ B(n - 1, n) = \int_0^{1}t^{n - 1}(1 - t)^{n}dt = \int_0^{1}\sum_{m = 0}^{n}\binom{n}{m}(-1)^{m}t^{n + m - 1}dt $ $ = \sum_{m = 0}^{n}\frac{(-1)^{m}}{n + m}\binom{n}{m} $ as desired. Now, returning to the original problem, note that $ \sum_{k = 0}^{n}(-1)^k(n - k)!(n + k)! = $ \[ (2n + 1)!\sum_{k = 0}^{n}(-1)^kB(n + k, n - k) = \] \[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}t^{n + k}(1 - t)^{n - k}dt = \] \[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\int_{0}^{1}\sum_{j = 0}^{n - k}\binom{n - k}{j}(-1)^jt^{n + k +j} dt = \] \[ (2n + 1)!\sum_{k = 0}^{n}(-1)^k\sum_{j = 0}^{n - k}\frac{(-1)^j}{n + k + j + 1}\binom{n - k}{j} = \] \[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\sum_{j + k = m}\binom{n - k}{j} = \] \[ (2n + 1)!\sum_{m = 0}^{n}\frac{(-1)^m}{n + m + 1}\binom{n + 1}{m} = \] \[ (2n + 1)!\left(B(n, n + 1) + \frac{(-1)^n}{2(n + 1)}\right) = \] \[ (2n + 1)!\left(\frac{n!(n + 1)!}{(2n + 2)!} + \frac{(-1)^n}{2n + 2}\right) \] And so we are done. Essentially this problem reduced to two applications of the binomial theorem on the integrand of the Beta function.