I have a question about combination series. i don't understand that $$
\sum_{r=0}^n \frac{(-1)^r}{n+r+1} {n \choose r} 
= \frac{
\sqrt{\pi}~ 2^{-2n-1} n !
}{
\left(n  + \frac{1}{2} \right)!
}
$$
How to explain what is the left series become to right form? I calculated by wolfram and received it but I couldn't that how to transform from a combination formula in the left series to the useful tool(ex: factorial or double). so I don't know about concrete solution.
(sorry, I'm not good at English because I was born in korea. If you don't like this terrible sentence, Please forgive me...)
 A: To evaluate the LHS introduce
$$f(z)  = (-1)^n n! \frac{1}{z+n+1} \prod_{q=0}^n \frac{1}{z-q}.$$
This has the property that
$$\mathrm{Res}_{z=r} f(z)
= (-1)^n n! \frac{1}{r+n+1}
\prod_{q=0}^{r-1} \frac{1}{r-q}
\prod_{q=r+1}^n \frac{1}{r-q}
\\ = (-1)^n n! \frac{1}{r+n+1}
\frac{1}{r!}
\frac{(-1)^{n-r}}{(n-r)!}
= \frac{1}{r+n+1}
(-1)^r {n\choose r}.$$
As we seek to compute
$$\sum_{r=0}^n \frac{1}{r+n+1}
(-1)^r {n\choose r}$$
and residues sum to zero with the residue at infinity being zero by
inspection we get for our sum
$$-\mathrm{Res}_{z=-n-1} f(z)
= - (-1)^n n! \prod_{q=0}^n \frac{1}{-n-1-q}
= n! \prod_{q=0}^n \frac{1}{n+1+q}
\\ = \frac{n!\times n!}{(2n+1)!}.$$
On the other hand working with the RHS we have with the Legendre
duplication formula
$$\frac{\sqrt{\pi} 2^{-2n-1} n!}{(n+1/2)!}
= \frac{\sqrt{\pi} 2^{-2n-1} n!}{\Gamma(n+3/2)}
\\ = \frac{\sqrt{\pi} 2^{-2n-1} \times n! \times n!}
{\Gamma(n+1) \Gamma(n+3/2)}
\\ = \frac{\sqrt{\pi} 2^{-2n-1} \times n! \times n!}
{2^{1-2(n+1)} \sqrt{\pi} \Gamma(2n+2)}
= \frac{n!\times n!}{(2n+1)!}.$$
We see that the LHS and the RHS are identical as claimed.
A: 
We obtain following the hint of @ZAhmed:
\begin{align*}
\color{blue}{\sum_{r=0}^n}&\color{blue}{ \frac{(-1)^r}{n+r+1} \binom{n}{r}}\\
&=\sum_{r=0}^n(-1)^r\int_{0}^1 z^{n+r}\,dz\binom{n}{r}\tag{1}\\
&=\int_{0}^1z^n\sum_{r=0}^n\binom{n}{r}(-z)^r\,dz\\
&=\int_{0}^1z^n(1-z)^n\,dz\tag{2}\\
&=B(n+1,n+1)\tag{3}\\
&=\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+2)}\tag{4}\\
&=\frac{\sqrt{\pi}\Gamma(n+1)}{2^{2n+1}\Gamma\left(n+\frac{3}{2}\right)}\tag{5}\\
&\,\,\color{blue}{=\frac{\sqrt{\pi}\,n!}{2^{2n+1}\left(n+\frac{1}{2}\right)!}}\tag{6}
\end{align*}
and the claim follows.

Comment:

*

*In (1) we use $\frac{1}{q+1}=\int_0^1{z^q}\,dz$


*In (2) we apply the binomial theorem.


*in (3) we observe we have the beta function.


*In (4) we represent the beta funtion with Gamma functions.


*In (5) we recall the Legendre duplication formula
\begin{align*}
\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z)
\end{align*}
evaluated at $z=n+1$.


*In (6) we use the identity $\Gamma(z+1)=z!$.
