Is the sum $\sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$ always the reciprocal of an integer $\big(\frac{(2n+1)!}{(n!)^2}\big)$? Denote the sum $$S_n :=  \sum_{k=0}^n{n \choose k}\frac{(-1)^{n-k}}{n+k+1}$$
This value arose in some calculations of polynomial coefficients. I'm not used to dealing with expressions of this sort. Does this arise anywhere else?
The first few values are: $1, \frac{-1}{6}, \frac{1}{30}, \frac{-1}{140}, \frac{1}{630}, \frac{-1}{2772}, \frac{1}{12012}$.
My first thoughts were Bernoulli numbers and primorials, but those seem to be red herrings. It looks like they are always the reciprocal of an integer, but I don't know how to prove that.
From OEIS A002457, the absolute values of the reciprocals appear to coincide with
$$a(n) = \frac{(2n+1)!}{(n!)^2}$$
If this is the case, the same page would imply that
$$S_n = \int_0^1(x(1-x))^ndx$$
I would really appreciate help with:

*

*Prove each $\frac{1}{S_n}$ is an integer.

*Is it actually the $a(n)$ above? If not, is there a nice description?

*Prove that the sign of $S_n$ alternates. (This is least important)

 A: We have
\begin{align*}
\int_{0}^{1} (x (1 - x))^n \ \text{dx}
& = \int_{0}^{1} (x - x^2)^n \ \text{dx} \\
& = \int_{0}^{1} \sum_{k = 0}^{n} \binom{n}{k} x^k (- x^2)^{n - k} \ \text{dx} \\
& = \sum_{k = 0}^{n} \binom{n}{k} (-1)^{n - k} \int_{0}^{1} x^{2 n - k} \ \text{dx} \\
& = \sum_{k = 0}^{n} \binom{n}{k} \frac{(-1)^{n - k}}{2 n - k + 1} \\
& \overset{\star}{=} \sum_{k = 0}^{n} \binom{n}{n - k} \frac{(-1)^k}{2 n - (n - k) + 1} \\
& = \sum_{k = 0}^{n} \binom{n}{k} \frac{(-1)^k}{n + k + 1}.
\end{align*}
where in step $(\star)$ we use the substitution $k \rightarrow n - k$ (thanks @RobPratt for the suggestion).
A: To evaluate the sum
$$\sum_{k=0}^n {n\choose k} \frac{(-1)^{n-k}}{n+k+1}$$
introduce the function
$$f(z) = \frac{n!}{n+1+z}
\prod_{q=0}^n \frac{1}{z-q}$$
which has the property that for $0\le k\le n$
$$\mathrm{Res}_{z=k} f(z)
= \frac{n!}{n+1+k}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = \frac{n!}{n+1+k}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^{n-k}}{n+1+k}.$$
Now with residues summing to zero and the residue at infinity being
zero by inspection we get for our sum the value
$$-\mathrm{Res}_{z=-n-1} f(z)
= - n! \prod_{q=0}^n \frac{1}{-n-1-q}
\\ = n! (-1)^{n} \prod_{q=0}^n \frac{1}{n+1+q}
= (-1)^{n} n! \frac{n!}{(2n+1)!}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{
\frac{(-1)^n}{2n+1} {2n\choose n}^{-1}.}$$
