I am tring to find the closed from of the following function: $$f(k) = \lim_{n \rightarrow \infty}\prod^{kn}_{i = 0}\frac{n-2i+3}{n-2i+2},$$ where $k \in [0,\frac{1}{2}]$
If the numerator is $n-2i+4$ instead of $n-2i+3$, the products can be nicely canceled to $\lim_{n \rightarrow \infty} \frac{n+3}{n-2kn+2} = \frac{1}{1-2k}$. But the relaxzation is rather big and I don't really know where to look for dealing with this product.
An "automatic" approach is to use the gamma function limit $\lim\limits_{x\to+\infty}\frac{\Gamma(x+a)}{x^a\ \Gamma(x)}=1$: since $$\prod_{i=0}^m\frac{n-2i+3}{n-2i+2}=\frac{\Gamma\left(\frac{n+5}2\right)\Gamma\left(\frac{n+2}2-m\right)}{\Gamma\left(\frac{n+4}2\right)\Gamma\left(\frac{n+3}2-m\right)},$$ we see that $$f(k)=\lim_{n\to\infty}\frac{(n/2)^{5/2}(n/2-kn)^{2/2}}{(n/2)^{4/2}(n/2-kn)^{3/2}}=(1-2k)^{-1/2}.$$