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 $n > m$: $$u_{m,n}(x) = (2n+1)\sin\left(\frac{\pi x}{2n+1}\right) \prod_{k=1}^{m}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right)$$ and $$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$ Show that the sequences, indexed by $n$, $\left(u_{m,n}(x)\right)_{n > m}$ and $\left(v_{m,n}(x)\right)_{n > m}$ are convergent in $\mathbb{R}^*$.
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 $n > m$:
$$u_{m,n}(x) = (2n+1)\sin\left(\frac{\pi x}{2n+1}\right) \prod_{k=1}^{m}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right)$$
and
$$v_{m,n}(x) = \prod_{k=m+1}^{n}\left(1 - \frac{\sin^2\left(\frac{\pi x}{2n+1}\right)}{\sin^2\left(\frac{k\pi}{2n+1}\right)}\right).$$
Show that the sequences, indexed by $n$, $\left(u_{m,n}(x)\right)_{n > m}$ and $\left(v_{m,n}(x)\right)_{n > m}$ are convergent in $\mathbb{R}^*$.