The limit of the product of $\frac{n}{n-k}$ where $k$ is constant and $n\rightarrow\infty$

Question

How do I calculate the limit of products such as: $$\lim_{n \to \infty} \frac{\prod_{x=k+1}^{n}x}{(n+1)\prod_{x=k+1}^{n}(x-k)}$$

First of all, I am assuming this is independent of k, since k is a constant. The ratio $\frac{n}{n-k}$ will go to $1$, and all the terms are finite but positive, so it seems that the limit should go to zero because $$\lim_{n \to \infty} \frac{\prod_{x=k+1}^{n}x}{\prod_{x=k+1}^{n}(x-k)}$$ will go to $\infty$ but it will grow less than $n$.

How do I make this more formal?

Answer 1

I believe you're asked to calculate $$\lim_{n\to\infty} \frac{n!/k!}{n(n-k)!} = \frac{1}{k!}\lim_{n\to\infty} \frac{(n-1)!}{(n-k)!}.$$ If $k=1$ the limit equals $1$. If $k=0$, it becomes $$ \lim_{n\to\infty}\frac{1}{n}=0. $$ If $k\geq 2$ then you have $$\lim_{n\to\infty}\frac{1}{k!}(n-1)(n-2)...(n-k+1)=+\infty.$$

Answer 2

Using the gamma function $$ \frac{\prod_{x=k+1}^{n}x}{(n+1)\prod_{x=k+1}^{n}(x-k)}=\frac{1}{\Gamma (k+1)}\,\,\,\frac{\Gamma (n+1)}{(n+1)\,\,\, \Gamma (n+1-k)}$$ If $k$ is fixed and $n\to \infty$ consider $$A=\log \left(\frac{\Gamma (n+1)}{(n+1)\,\,\, \Gamma (n+1-k)}\right)$$ and use Stirling approximation twice and continue with Taylor series to obtain $$A=(k-1)\log(n)-\frac{k^2-k+2}{2 n}+O\left(\frac{1}{n^2}\right)$$ that is to say $$\frac{\Gamma (n+1)}{(n+1)\,\,\, \Gamma (n+1-k)}=e^A=n^{k-1} \left( 1-\frac{k^2-k+2}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$ $$ \frac{\prod_{x=k+1}^{n}x}{(n+1)\prod_{x=k+1}^{n}(x-k)}\sim\frac{n^{k-1}}{\Gamma (k+1)}$$ Now, consider the cases