<s>[INST] Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$? It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or it diverges?
 [/INST] Calling $\Pi$ the partial product up to $N$, for $0<|x|<\pi/\sqrt{3}$ we have
$$ \begin{array}{ll} 
-\ln\Pi & =\displaystyle \sum_{n=1}^N \ln\Bigg(\frac{1}{1-\frac{(2n+1)x^2}{n^2\pi^2}}\Bigg) \\[5pt] & > \displaystyle \sum_{n=1}^N \ln\Big(1+\frac{(2n+1)x^2}{n^2\pi^2}\Big) \\[5pt]
& >\displaystyle \sum_{n=1}^N \left[\frac{(2n+1)x^2}{n^2\pi^2}-\frac{1}{2}\left(\frac{(2n+1)x^2}{n^2\pi^2}\right)^2 \right] \\[5pt]
& >\displaystyle \left[\frac{2x^2}{\pi^2}\sum_{n=1}^N \frac{1}{n}\right] - \mathcal{O}(1) ~~~\to\infty
\end{array} $$
Thus, $\ln \Pi\to-\infty$ so $\Pi\to0^+$ as $N\to\infty$.
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