Properties of the Gamma function

Question

How I can show that

$$\prod_{r=2}^{N}\frac{\Gamma^2(r\alpha+1)}{\Gamma((r-1)\alpha+1)\Gamma((r+1)\alpha+1)} \lt 1$$

for all $N \gt 2$ and $0 \lt α\leq 1$?

Thanks

Answer 1

Notice that the product telescopes to

$$ \frac{\Gamma(2\alpha+1)\Gamma(N\alpha+1)}{\Gamma(\alpha+1)\Gamma((N+1)\alpha+1)}.$$

Fleshing out the comment by andre, let us use the sum formula for the gamma function,

$$\Gamma(x+y)=\frac{\Gamma(x)\Gamma(y)}{B(x,y)},$$

where $B(x,y)=\int_0^1 t^{x-1} (1-t)^{y-1} dt$ is the beta function.

Thus $$\Gamma(2\alpha+1)=\Gamma(\alpha+1)\Gamma(\alpha)/B(\alpha+1,\alpha)$$ and $$\Gamma((N+1)\alpha+1)=\Gamma(N\alpha+1)\Gamma(\alpha)/B(N\alpha+1,\alpha).$$ Plugging this into the above, our claim becomes

$$\frac{B(N\alpha+1,\alpha)}{B(\alpha+1,\alpha)}<1.$$

Recalling the definition of the beta function, the claim is simply

$$\int_0^1 t^{N\alpha} (1-t)^{\alpha-1} dt < \int_0^1 t^{\alpha} (1-t)^{\alpha-1} dt.$$

But $0\le t^\alpha\le 1$, so since $N>1$ we have $(t^\alpha)^N< t^\alpha$ for all $0<t<1$, so this one is obvious.