Find the value of the Infinite product in terms of k which is a positive integer

Question

$$\prod_{n=k+1}^{+\infty}\left(1-\frac{k^2}{n^2}\right)$$

The only help we have been able to find is that of Euler, anything would be amazing!

Answer 1

You are looking for: $$\begin{eqnarray*} \prod_{n\geq 1}\left(1-\frac{k^2}{(n+k)^2}\right)&=&\lim_{m\to +\infty}\frac{m! (2k+1)(2k+2)\cdot\ldots\cdot(2k+m)}{(k+1)^2 (k+2)^2\cdot\ldots\cdot(k+m)^2}\\&=&\lim_{m\to+\infty}\frac{m!(2k+m)!k!^2}{(2k)!(k+m)!^2}\\&=&\frac{1}{\binom{2k}{k}}\lim_{m\to +\infty}\frac{m!(2k+m)!}{(k+m)!^2}\\&=&\frac{1}{\binom{2k}{k}}\lim_{m\to +\infty}\frac{(m+k+1)(m+k+2)\cdot\ldots\cdot(m+k+k)}{(m+1)(m+2)\cdot\ldots\cdot(m+k)}\end{eqnarray*} $$ where the last limit equals $1$, since $\lim_{m\to +\infty}\frac{m+k+j}{m+j}=1$ and we are multiplying $k$ (that is fixed) of these limits. It follows that:

$$\prod_{n\geq k+1}\left(1-\frac{k^2}{n^2}\right) = \color{red}{\frac{1}{\binom{2k}{k}}}.$$