An equality between a product and a combinatorial sum I'm trying to prove the following identity (of which I numerically verified the truth) :
$$\text{For every $n\in\mathbb{N}^*$ and $\alpha \in \mathbb{R}\setminus\lbrace-2k\text{ }|\text{ }k\in\mathbb{N}\rbrace$,}$$
$$\text{$\prod\limits_{k=1}^{n}$}\left[1-\frac{1}{2k+\alpha}\right]=\frac{\alpha}{4^{n}}{2n \choose n}\text{$\sum\limits_{k=0}^{n}$}\frac{{n \choose k}^2}{{2n \choose 2k}}\frac{1}{2k+\alpha}$$
I've tried induction, unsuccessfully. Tbh, I don't really have any other ideas for tackling it. Products such as this are not that easy to work with.
Any ideas or suggestion ?
 A: This is a mere supplement to @MarkoRiedels answer. Mimicking his solution with a slight change we can simplify the proof.

We start with OP's left-hand side and obtain
  \begin{align*}
\color{blue}{\prod_{k=1}^n}&\color{blue}{\left(1-\frac{1}{2k+\alpha}\right)}\\
&=\prod_{k=1}^n\frac{2k-1+\alpha}{2k+\alpha}\\
&=\alpha\prod_{k=1}^n(2k-1+\alpha)\prod_{k=0}^n\frac{1}{2k+\alpha}\tag{1}\\
&=\alpha\sum_{k=0}^n\frac{1}{2k+\alpha}\mathrm{Res}_{\alpha=-2k}\prod_{q=1}^n(2q-1+\alpha)\prod_{q=0}^n\frac{1}{2q+\alpha}\\
&=\alpha\sum_{k=0}^n\frac{1}{2k+\alpha}\prod_{q=1}^n(2q-1-2k)\prod_{q=0}^{k-1}\frac{1}{2q-2k}\prod_{q=k+1}^n\frac{1}{2q-2k}\\
&=\frac{\alpha}{2^n}\sum_{k=0}^n\frac{1}{2k+\alpha}(-1)^k\prod_{q=1}^k(2k+1-2q)\prod_{q=k+1}^n(2q-1-2k)\\
&\qquad\qquad\cdot\prod_{q=0}^{k-1}\frac{1}{q-k}\prod_{q=k+1}\frac{1}{q-k}\\
&=\frac{\alpha}{2^n}\sum_{k=0}^n\frac{1}{2k+\alpha}(-1)^k(2k-1)!!(2n-1-2k)!!\cdot\frac{1}{(-1)^kk!}\cdot\frac{1}{(n-k)!}\\
&=\frac{\alpha}{2^n}\sum_{k=0}^n\frac{1}{2k+\alpha}\cdot\frac{(2k)!}{2^kk!}\cdot\frac{(2n-2k)!}{2^{n-k}(n-k)!}\cdot\frac{1}{(-1)^kk!}\cdot\frac{1}{(n-k)!}\\
&=\frac{\alpha}{4^n}\binom{2n}{n}\sum_{k=0}^n\frac{1}{2k+\alpha}\cdot\frac{n!}{k!(n-k)!}\cdot\frac{n!}{k!(n-k)!}\cdot\frac{(2k)!(2n-2k)!}{(2n)!}\\
&\,\,\color{blue}{=\frac{\alpha}{4^n}\binom{2n}{n}\sum_{k=0}^n\binom{n}{k}^2\binom{2n}{2k}^{-1}\frac{1}{2k+\alpha}}
\end{align*}
and the claim follows.

Comment:


*

*In (1) is the slight difference. Instead of factoring out $1+\alpha$ we multiply with $\frac{\alpha}{\alpha}$.

