Infinite product made rigorous I was reading a proof in Koblitz's book "p-adic numbers, p-adic analysis and zeta-functions" in page 22 that says:
For all real numbers $x$, the infinite product 
$$\pi x \prod_{n=1}^\infty \left(1+\dfrac{x^2}{n^2} \right)$$
converges and equals $\sinh(\pi x)$.
Then, at some point the author states this (where $n=2k+1$):
$$\dfrac{\sin(\pi x)}{n\sin(\pi x/n)}=\prod_{r=1}^k \left( 1- \dfrac{\sin^2(\pi x/n)}{\sin^2(\pi r/n)} \right)$$
Now take the limit of both sides as $n = 2k + 1 \rightarrow \infty$. The left-hand side approaches $\sin(\pi x)/\pi x$. For $r$ small relative to $n$ the $r$th term in the product approaches $1-((\pi x/n)/(\pi r/n))^2 = 1 - (x^2/r^2)$. It then follows that the product converges to
$$\prod_{r=1}^\infty (1-(x^2/r^2))$$
Then it says that the rigorous justification of the previous statement is straightforward, and is left as an exercise.
The proposed exercise is proving that
$$\prod_{r=1}^k \dfrac {1- \frac{\sin^2(\pi x/(2k+1))}{\sin^2(\pi r/(2k+1))}}{1-\frac{x^2}{r^2}}$$
converges to $1$ as $k$ approaches infinity.
I know that, since the numerator of the fraction converges (to $\frac{\sin(\pi x)}{\pi x}$ as said before) showing this limit is $1$, proves that the denominator limit of products converges to the same value but I don't know how to rigurously prove this limit since it involves $k$ as the value por the upper limit in the product and it also appears in the factors.
Could someone explain how to prove this?
Thanks a lot in advance.
 A: Define 
$$f_r(k) = \begin{cases} - \frac{\sin^2(\pi x/(2k+1))}{\sin^2(\pi r/(2k+1))}, &r \leqslant k\\ 0, &r > k \end{cases}$$
Using the inequality $2z/\pi \leqslant \sin z \leqslant z$, we can show that $|f_r(k)| \leqslant \frac{\pi^2x^2}{4r^2}$.  Hence, the series $\sum_{r=1}^\infty |f_r(k)|$ is uniformly convergent for all $k \in \mathbb{N}$ and, consequently the product $\prod_{r=1}^\infty (1 +f_r(k))$ is also uniformly convergent.
Thus, we can interchange limit and product to obtain
$$\lim_{k \to \infty}\prod_{r=1}^k \left[ 1 - \frac{\sin^2(\pi x/(2k+1))}{\sin^2(\pi r/(2k+1))}\right] = \lim_{k \to \infty}\prod_{r=1}^\infty (1 + f_r(k)) = \prod_{r=1}^\infty(1 + \lim_{k \to \infty}f_r(k))$$
Finally, we have
$$\lim_{k \to \infty} f_r(k) = \lim_{k \to \infty} -\frac{\sin^2(\pi x/(2k+1))}{\sin^2(\pi r/(2k+1))} =  \lim_{k \to \infty}- \frac{x^2}{r^2} \frac{\sin^2(\pi x/(2k+1))}{(\pi x/(2k+1))^2}\frac{(\pi r/(2k+1))^2}{\sin^2(\pi r/(2k+1))} \\ = - \frac{x^2}{r^2},$$
and, therefore,
$$\lim_{k \to \infty}\prod_{r=1}^k \left( 1 - \frac{\sin^2(\pi x/(2k+1))}{\sin^2(\pi r/(2k+1))}\right) = \prod_{r=1}^\infty \left(1 - \frac{x^2}{r^2} \right)$$
