How do I show this :$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$

Question

let $a, b , k$ integes with $a\neq 0$, The bellow formula is the product of arithmitic progression , i want to know if there any proof show that :

$$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$

?

Answer 1

Note that\begin{align}\Gamma\left(n+1+\frac ba\right)&=\Gamma\left(n+\frac ba\right)\times\left(n+\frac ba\right)\\&=a^{-1}\Gamma\left(n+\frac ba\right)\times\left(an+b\right).\end{align}Using this equality together with induction, you'll get the equality that you want to prove.

Answer 2

Take $g(k)=u^{k}\Gamma(\frac{s}{u}+k)$,

we will use the fundamental property of the gamma function $$\Gamma(\frac{s}{u}+k+1)=\Gamma(\frac{s}{u}+k)\left(\frac{s}{u}+k\right)$$

And the telescopic product $$\prod^n_{k=1} \frac{g(k+1)} {g(k)}=\frac{g(n+1)}{g(1)}. $$

We have

$$\frac{g(k+1)}{g(k)}=u^{k+1}\Gamma(\frac{s}{u}+k+1) \frac{1}{u^{k} }\frac{1 }{ \Gamma(\frac{s}{u}+k)}=u\frac{\Gamma(\frac{s}{u}+k) (\frac{s}{u}+k)}{\Gamma(\frac{s}{u}+k)}=u(\frac{s}{u}+k)=s+uk.$$

So $\frac{g(k+1)}{g(k)}=s+uk$, the product is telescopic. Apply $\prod\limits_{k=1}^n$ on both sides.

$$\prod_{k=1}^{n} (s+uk) =\frac{g(n+1)}{g(1)}=\frac{u^{n}\Gamma(\frac{s}{u}+n+1)}{\Gamma(\frac{s}{u}+1)} $$

Answer 3

Here is a way to show your result using the rising factorial $x^{(n)}$ which is defined as $$x^{(n)} = x(x + 1)(x+2) \cdots (x + n - 1), \quad n \in \mathbb{N}.$$

For your product, we can write \begin{align*} \prod_{k = 1}^n (ak + b) &= (a + b)(2a + b) \cdots (an + b)\\ &= a^n \left (1 + \frac{b}{a} \right ) \left (2 + \frac{b}{a} \right ) \cdots \left (n + \frac{b}{a} \right )\\ &= a^n \frac{\frac{b}{a} \left (\frac{b}{a} + 1 \right ) \left (\frac{b}{a} + 2 \right ) \cdots \left (\frac{b}{a} + n \right )}{b/a}\\ &= a^n \frac{\left (\frac{b}{a} \right )^{(n + 1)}}{b/a}.\tag1 \end{align*}

Now from properties for the rising factorial, as $$x^{(n)} = \frac{\Gamma (x + n)}{\Gamma (x)},$$ we can rewritten (1) as $$\prod_{k = 1}^n (ak + b) = a^n \frac{\Gamma \left (\frac{b}{a} + n + 1 \right )}{\frac{b}{a} \cdot \Gamma \left (\frac{b}{a} \right )} = a^n \frac{\Gamma \left (n + 1 + \frac{b}{a} \right )}{\Gamma \left (1 + \frac{b}{a} \right )},$$ as expected.