<s>[INST] Does $\prod_{m=1}^\infty \frac{1}{m^2}$ have a closed form? We know that $$\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6},$$ but what about the product of the reciprocal of the squares:
$$\prod_{m=1}^\infty \frac{1}{m^2}?$$
Do we use a different product representation to compute this? Maybe the cosine product 
$$\cos{x} = \prod_{m=1}^\infty \left(1-\frac{x^2}{\pi^2\left(m-\frac{1}{2}\right)^2}\right).$$
 [/INST] For a sum to converge, it is necessary that the terms converge to $0$ (the neutral element for the sum). $\frac{1}{n^2}$ satisfies this.
For a product to converge to a nonzero value, it is necessary that the factors converge to $1$ (the neutral element for the product). $\frac{1}{n^2}$ does not satisfies this, so the product diverges. In this case, the product is $0$.
But because $\frac{1}{n^2} \to 0$, we do of course have that
$$1+\frac{1}{n^2} \to 1$$
and
$$1-\frac{1}{n^2} \to 1$$ 
So perhaps more interesting product analogies of the sum you mention are
$$\begin{aligned}
\prod_{n=1}^\infty 1+\frac{1}{n^2} &=\frac{\sinh(\pi)}{\pi} = \frac{-e^{-\pi}}{2\pi} + \frac{e^\pi}{2\pi}\\
\prod_{n=2}^\infty 1-\frac{1}{n^2} &= \frac{1}{2}
\end{aligned}
$$
</s>