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 ?
Best Answer
For the identity
$$\bbox[5px,border:2px solid #00A000]{ \prod_{k=1}^n \left[ 1 - \frac{1}{2k+\alpha} \right] = \frac{\alpha}{4^n} {2n\choose n} \sum_{k=0}^n {n\choose k}^2 {2n\choose 2k}^{-1} \frac{1}{2k+\alpha}}$$
we work with $n\ge 2$ and we first simplify the RHS
$$\frac{\alpha}{4^n} \frac{(2n)!}{n!\times n!} \sum_{k=0}^n \frac{n!^2}{k!^2 (n-k)!^2} \frac{(2k)! (2n-2k)!}{(2n)!} \frac{1}{2k+\alpha} \\ = \frac{\alpha}{4^n} \sum_{k=0}^n {2k\choose k} {2n-2k\choose n-k} \frac{1}{2k+\alpha} \\ = \frac{1}{4^n} {2n\choose n} + \frac{\alpha}{4^n} \sum_{k=1}^n {2k\choose k} {2n-2k\choose n-k} \frac{1}{2k+\alpha} .$$
Now for the LHS we find
$$\prod_{k=1}^n \frac{2k-1+\alpha}{2k+\alpha} = \prod_{k=1}^n (2k-1+\alpha) \prod_{k=1}^n \frac{1}{2k+\alpha} \\ = (1+\alpha) \prod_{k=2}^n (2k-1+\alpha) \prod_{k=1}^n \frac{1}{2k+\alpha}.$$
We may apply partial fractions by residues to the two products (degree of numerator is less than degree of denominator and poles are all simple) for the variable $\alpha$ and get
$$(1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} \mathrm{Res}_{\alpha=-2k} \prod_{q=2}^n (2q-1+\alpha) \prod_{q=1}^n \frac{1}{2q+\alpha} \\ = (1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} \prod_{q=2}^n (2q-1-2k) \prod_{q=1}^{k-1} \frac{1}{2q-2k} \prod_{q=k+1}^{n} \frac{1}{2q-2k} \\ = \frac{1}{2^{n-1}} (1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} \prod_{q=2}^n (2q-1-2k) \prod_{q=1}^{k-1} \frac{1}{q-k} \prod_{q=k+1}^{n} \frac{1}{q-k} \\ = \frac{1}{2^{n-1}} (1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} \frac{(-1)^{k-1}}{(k-1)!} \frac{1}{(n-k)!} \prod_{q=2}^n (2q-1-2k).$$
For the remaining product we get
$$\prod_{q=2}^n (2q-1-2k) = \prod_{q=2}^k (2q-1-2k) \prod_{q=k+1}^n (2q-1-2k) \\ = \frac{(-1)^{k-1} (2k-2)!}{(k-1)! \times 2^{k-1}} \prod_{q=1}^{n-k} (2q-1) \\ = \frac{(-1)^{k-1} (2k-2)!}{(k-1)! \times 2^{k-1}} \frac{(2n-2k)!}{(n-k)! \times 2^{n-k}}.$$
Collecting everything we have for the partial fraction decomposition
$$\frac{1}{2^{n-1}} (1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} \frac{(-1)^{k-1}}{(k-1)!} \frac{1}{(n-k)!} \frac{(-1)^{k-1} (2k-2)!}{(k-1)! \times 2^{k-1}} \frac{(2n-2k)!}{(n-k)! \times 2^{n-k}}$$
which is
$$\bbox[5px,border:2px solid #00A000]{ \frac{1}{4^{n-1}} (1+\alpha) \sum_{k=1}^n \frac{1}{2k+\alpha} {2k-2\choose k-1} {2n-2k\choose n-k}.}$$
Next to conclude we extract the coefficient on $[\alpha^q]$ and show that the RHS and the LHS yield the same, proving the claim (no pole at $\alpha=0$). We get for $q=0$ that we must have
$$\frac{1}{4^n} {2n\choose n} = \frac{1}{4^{n-1}} \sum_{k=1}^n \frac{1}{2k} {2k-2\choose k-1} {2n-2k\choose n-k} \\ = \frac{1}{2^{2n-1}} \sum_{k=0}^{n-1} \frac{1}{k+1} {2k\choose k} {2n-2-2k\choose n-1-k}.$$
We recognize Catalan numbers with OGF
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}.$$
Continuing,
$$\frac{1}{2^{2n-1}} \sum_{k=0}^{n-1} \frac{1}{k+1} {2k\choose k} [z^{n-1-k}] (1+z)^{2n-2-2k} \\ = \frac{1}{2^{2n-1}} [z^{n-1}] \sum_{k=0}^{n-1} \frac{1}{k+1} {2k\choose k} z^k (1+z)^{2n-2-2k}.$$
Now when $k\gt n-1$ there is no contribution to the coefficient extractor and we continue with
$$\frac{1}{2^{2n-1}} [z^{n-1}] (1+z)^{2n-2} \sum_{k\ge 0} \frac{1}{k+1} {2k\choose k} z^k (1+z)^{-2k} \\ = \frac{1}{2^{2n-1}} [z^{n-1}] (1+z)^{2n-2} \frac{1-\sqrt{1-4z/(1+z)^2}}{2z/(1+z)^2} \\ = \frac{1}{2^{2n}} [z^{n}] (1+z)^{2n-2} \frac{1-\sqrt{1-4z/(1+z)^2}}{1/(1+z)^2} \\ = \frac{1}{2^{2n}} [z^{n}] (1+z)^{2n-1} (1+z-\sqrt{1-2z+z^2}) \\ = \frac{1}{2^{2n}} [z^{n}] (1+z)^{2n-1} 2z = \frac{1}{2^{2n}} [z^{n-1}] 2 (1+z)^{2n-1} = \frac{1}{2^{2n}} 2 {2n-1\choose n-1} \\ = \frac{1}{2^{2n}} {2n\choose n}.$$
We have equality for $q=0.$ For $q\ge 1$ we must have that
$$\frac{1}{4^n} \sum_{k=1}^n {2k\choose k} {2n-2k\choose n-k} [\alpha^{q-1}] \frac{1}{2k} \frac{1}{1+\alpha/2/k} \\ = \frac{1}{4^n} \sum_{k=1}^n {2k\choose k} {2n-2k\choose n-k} (-1)^{q-1} \frac{1}{2^qk^q}$$
is the same as
$$\frac{1}{4^{n-1}} \sum_{k=1}^n ([\alpha^q] + [\alpha^{q-1}]) \frac{1}{2k+\alpha} {2k-2\choose k-1} {2n-2k\choose n-k}.$$
This is
$$\frac{1}{4^{n-1}} \sum_{k=1}^n \left((-1)^q \frac{1}{2^{q+1} k^{q+1}} + (-1)^{q-1}\frac{1}{2^qk^q}\right) {2k-2\choose k-1} {2n-2k\choose n-k} \\ = \frac{1}{4^{n-1}} \sum_{k=1}^n \left(-\frac{1}{2k} + 1\right) (-1)^{q-1}\frac{1}{2^qk^q} {2k-2\choose k-1} {2n-2k\choose n-k} \\ = \frac{1}{2^{2n-2}} \sum_{k=1}^n \frac{2k-1}{2k} (-1)^{q-1}\frac{1}{2^qk^q} \frac{k}{2k-1} {2k-1\choose k} {2n-2k\choose n-k} \\ = \frac{1}{2^{2n-1}} \sum_{k=1}^n (-1)^{q-1}\frac{1}{2^qk^q} {2k-1\choose k-1} {2n-2k\choose n-k} \\ = \frac{1}{2^{2n-1}} \sum_{k=1}^n (-1)^{q-1}\frac{1}{2^qk^q} \frac{k}{2k} {2k\choose k} {2n-2k\choose n-k} \\ = \frac{1}{4^n} \sum_{k=1}^n (-1)^{q-1}\frac{1}{2^qk^q} {2k\choose k} {2n-2k\choose n-k}.$$
We again have equality. This concludes the proof.