Finding recurrence relationship involving alternating coefficients

Question

I'm trying to express the following recursive relationship as a summation \begin{align} a_2 &= \frac{-\alpha(\alpha+1)}{2!}a_0 \\ a_4 &= \frac{\alpha(\alpha+1)(\alpha-2)(\alpha+3)}{4!}a_0 \\ a_6 &= \frac{-\alpha(\alpha+1)(\alpha-2)(\alpha+3)(\alpha-4)(\alpha+5)}{6!}a_0\\ & \ \ \vdots \end{align}

Attempt

Genouinely confused , I suppose it should look something along those lines $$ a_{2n} = \sum_{n\ge0}\frac{(-1)^n (\alpha+1)\dots(\alpha +(2n-1))}{(2n)!}a_0,$$ but I'm not sure if something should go in between the two factors in the numerator.

Answer 1

Alternating signs could usually be represented by powers of $-1$:

$$a_{2n}=\sum_{n\ge0}\frac{(-1)^n\prod_{i=0}^{2n-1}\left(\alpha-(-1)^i\cdot i\right)}{(2n)!}a_0$$

$$=\sum_{n\ge0}\frac{(-1)^n\prod_{i=0}^{2n-1}\left(\alpha+(-1)^{i+1}\cdot i\right)}{(2n)!}a_0,$$

and alternating series could usually be splitted into several series:

$$a_{2n}=\sum_{n\ge0}\frac{(-1)^n\prod_{i=0}^{n-1}(\alpha-2i)\prod_{i=0}^{n-1}(\alpha+(2i+1))}{(2n)!}a_0,$$

$$=\sum_{n\ge0}\frac{(-1)^n\ \alpha(\alpha-2)...(\alpha-2(n-1))\ \cdot\ (\alpha+1)(\alpha+3)...(\alpha+(2n-1))}{(2n)!}a_0.$$

Answer 2

To handle odd as well as even $a_n$, you could write, with $c$ for $\alpha$, $a_n =a_0\dfrac{(-1)^{\lfloor n/2 \rfloor}\prod_{k=0}^{n-1} (c+(-1)^{k+1}k)}{n!} $.

Then

$\begin{array}\\ a_{2n} &=a_0\dfrac{(-1)^{n}\prod_{k=0}^{2n-1} (c+(-1)^{k+1}k)}{(2n)!}\\ &=a_0\dfrac{(-1)^{n}}{(2n)!}\prod_{k=0}^{2n-1} (c+(-1)^{k+1}k)\\ &=a_0\dfrac{(-1)^{n}}{(2n)!}\prod_{k=0}^{n-1} (c-2k)(c+2k+1)\\ &=a_0\dfrac{(-1)^{n}}{(2n)!}\prod_{k=0}^{n-1} (c-2k)\prod_{k=0}^{n-1}(c+2k+1)\\ &=a_0\dfrac{(-1)^{n}}{(2n)!}2^n\prod_{k=0}^{n-1} (c/2-k)2^n\prod_{k=0}^{n-1}((c+1)/2+k)\\ &=a_0\dfrac{(-1)^{n}4^n}{(2n)!}\dfrac{\Gamma(c/2+1)\Gamma((c+2n+1)/2)}{\Gamma(c/2-n+1)\Gamma((c+1)/2)}\\ \text{and}\\ \dfrac{a_{2n+2}}{a_n} &=- \dfrac{(c-2n)(c+2n+1)}{(2n+2)(2n+1)}\\ \end{array} $

If you know that $c$ is an integer, then other simplifications can be made.