Let $x \in \mathbb{R} \backslash \mathbb{Z}$. Let $m \in \mathbb{N}$ such that $m > |x|$. We set, for $n \in \mathbb{N}$ such that $n > m$: $$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$ We denote $v_m(x)$ the limit of $\left(v_{m,n}(x)\right)_{n > m}$. Show that, for $n \in \mathbb{N}$ such that $n > m$, we have $$1 \geqslant v_{m,n}(x) \geqslant \prod_{k=m+1}^{n}\left(1 - \frac{\pi^2 x^2}{4k^2}\right)$$ and deduce that $\lim_{m \rightarrow +\infty} v_m(x) = 1$.
Let $x \in \mathbb{R} \backslash \mathbb{Z}$. Let $m \in \mathbb{N}$ such that $m > |x|$. We set, for $n \in \mathbb{N}$ such that $n > m$:
$$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$
We denote $v_m(x)$ the limit of $\left(v_{m,n}(x)\right)_{n > m}$.
Show that, for $n \in \mathbb{N}$ such that $n > m$, we have
$$1 \geqslant v_{m,n}(x) \geqslant \prod_{k=m+1}^{n}\left(1 - \frac{\pi^2 x^2}{4k^2}\right)$$
and deduce that $\lim_{m \rightarrow +\infty} v_m(x) = 1$.