I'm trying find an alternate (shorter?) expression for product of $\,\dfrac{x^2-1}{x^2}\,$ over integers greater than $\,1\,$, i.e.$\,\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdots\quad$ An integral is a sum that would diverge, not a product, and we can see by inspection that this converges to $\,1.\quad$ How can this product be expressed differently?
Update: An answer and a comment indicate convergence to $\,\dfrac{1}{2}\,$ but it appears to me that they are multiplying zero and infinity which seems to me, undefined.