I have a question about combination series. i don't understand that

Question

$$ \sum_{r=0}^n \frac{(-1)^r}{n+r+1} {n \choose r} = \frac{ \sqrt{\pi}~ 2^{-2n-1} n ! }{ \left(n + \frac{1}{2} \right)! } $$

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...)

Answer 1

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.

Answer 2

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 $\Gamma(z+1)=z!$.