Rewriting $B(n+1/2,l+1)$ using factorials of integers

Question

I was trying to prove: $$ \mathrm{B}(n+\frac12,l+1)=2^{2l+1}l!\frac{(2n)!}{n!}\frac{(l+n)!}{(2l+2n+1)!} $$ where $n$ and $l$ are both positive integers and $\mathrm{B}$ is the beta function.

I tried: $$ \begin{aligned} \mathrm{B}(n+\frac12,l+1)&=\frac{\Gamma(n+1/2)\Gamma(l+1)}{\Gamma(n+l+3/2)} \\ &=l!\frac{\Gamma(1/2)\prod^n_{k=1}(k-1/2)}{\Gamma(1/2)\prod^{n+l+1}_{k=1}(k-1/2)} \\ &=l!\prod^{n+l+1}_{k=n+1}(k-\frac12)^{-1} \end{aligned} $$ but I don't know what to do next. Please correct me if I have made any stupid mistake.

Answer 1

HINT

Rewrite your expression as $$\prod_{k=n+1}^{n+l+1}\frac1{k-\frac12}=\prod_{k=n+1}^{n+l+1}\frac{2}{2k-1}$$ And then use the doubel factorial, or equivalently the fact that $$(2n-1)(2n-3)\cdots(3)(1)=\frac{(2n)(2n-1)\cdots(2)(1)}{2^n\cdot(n)(n-1)\cdots(2)(1)}$$ Suitable index considerations yield the result.

Answer 2

Starting from @mrtaurho's solution, using Pochhammer's symbols $$\prod_{k=n+1}^{n+l+1}\frac{2}{2k-1}=\frac{1}{\left(n+\frac{1}{2}\right)_{l+1}}$$ which can also write $$\frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma \left(n+l+\frac{3}{2}\right)}$$