# Proof that these two expressions are equivalent

I'm looking for algebraic proof that

$$y=\lim_{N \to \infty}\frac{\prod\limits_{n=0}^{N}x-n}{\prod\limits_{n=0}^{N-\lfloor x\rfloor}{x-\lfloor x\rfloor-n}}=\frac{\Gamma{(x+1)}}{\Gamma{(x-\lfloor x\rfloor}+1)}$$

I know that they are equivalent because when I graph $y$ vs. $x$ for both of these, the functions intersect at every point.

I can see how they are related when $x$ is an integer. Although the middle expression is undefined for those values (it requires dividing by zero), taking the limit as $x$ goes to $z$ ($\lim{x \to z}$, where $z \in \Bbb Z$) will make both functions equal to $z$ factorial.

I'm not sure about values in which $x$ is not an integer though, which is why I'm asking this question.

The reason I want to prove that this equation is true, is because I think that if you can find a good approximation for $\Gamma{(x-\lfloor x\rfloor+1)}$ ($-\frac{1}{9}|\sin \pi x \space|+1$ for example) then you could also approximate the gamma function pretty well.

$\small\text{Graph of the function}$

$\small\text{Graph of} \space \space \Gamma{(x-\lfloor x\rfloor+1)} \space\space\text{and} \space -\frac{1}{9}\left| \sin \left( \pi x \right) \right|+0.75$

• Use the defining relation $\Gamma(x+1) = x\Gamma(x)$, $\lfloor x\rfloor$ times on the right hand side to $\Gamma(1+x) = x(x-1)\cdots(x-\lfloor x\rfloor+1)\Gamma(x-\lfloor x\rfloor+1)$. For a formal proof you can use induction. Then show that the left hand side also factors into $x(x-1)\cdots(x-\lfloor x\rfloor+1)$. – Winther Mar 21 '16 at 1:54
• be careful, it is obvious that $\displaystyle \frac{\prod_{n=0}^\infty (x-n)}{\prod_{n=0}^\infty (x- \lfloor x \rfloor - n)}$ doesn't exists ( it is $\frac{\infty}{\infty}$). hence, what you meant was $\displaystyle \prod_{n=0}^\infty \frac{x-n}{x- \lfloor x \rfloor - n} = \lim_{N \to \infty} \prod_{n=0}^N \frac{x-n}{x- \lfloor x \rfloor - n} = \lim_{N \to \infty} \frac{\prod_{n=0}^N (x-n)}{\prod_{n=0}^N x- \lfloor x \rfloor - n}$ which is not the same as $\displaystyle \frac{\lim_{N \to \infty} \prod_{n=0}^N (x-n)}{\lim_{N \to \infty} \prod_{n=0}^N x- \lfloor x \rfloor - n}$ – reuns Mar 21 '16 at 2:08
• Ok thanks for pointing that out @user1952009 – user311559 Mar 21 '16 at 2:16
• @Winther I've thought of doing that, but I think it would only work as a proof for $x>0$ – user311559 Mar 22 '16 at 22:57

We'll start by looking at one part of the limit you give, namely, $$\prod_{n=0}^{N}x-n$$ which, expanded, gives $$x(x-1)(x-2)\ldots(x-(N-1))(x-N)$$ assuming $N$ is a non-negative integer. This is equal to $$\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-N\right)}$$ which can be shown using $\Gamma(a+1)=a\Gamma(a)$ for all $a$.
A similar manipulation can be done with the denominator of the limit you gave, substituting $N$ for $N-\lfloor x\rfloor$ and $x$ for $x - \lfloor x \rfloor$, giving $$\frac{\Gamma(x-\lfloor x\rfloor+1)}{\Gamma(x-N)}$$ after a small amount of simplification. This is only true if $N-\lfloor x\rfloor$ is a non-negative integer, as before.
Together these give $$\lim_{N\rightarrow \infty}\frac{\frac{\Gamma\left(x+1\right)}{\Gamma\left(x-N\right)}}{\frac{\Gamma(x-\lfloor x\rfloor+1)}{\Gamma(x-N)}} = \frac{\Gamma\left(x+1\right)}{\Gamma(x-\lfloor x\rfloor+1)}$$
The reason the limit cannot be dropped in the original expression—despite the cancellation here—is due to the fact that $N-\lfloor x\rfloor$ must be a non-negative integer for a product to be taken.