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 
$$\cos{x} = \prod_{m=1}^\infty \left(1-\frac{x^2}{\pi^2\left(m-\frac{1}{2}\right)^2}\right).$$
 A: $$\prod_{m=1}^n \frac{1}{m^2}=\frac1{(n!)^2},$$ which tends to $0$ at the speed of light.
A: Note that one clearly has
$$0 \le \prod_{m=1}^N\dfrac{1}{m^2} \le \dfrac{1}{N^2}.$$
The RHS tends to $0$ as $N \to \infty$ which gives us that the sequence of partial products converges to $0$. (Which is the same as saying that the product diverges (!) to $0$.)

Thus, to answer your question: Yes, the product does have a closed form!
A: 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
$$\begin{aligned}
\prod_{n=1}^\infty 1+\frac{1}{n^2} &=\frac{\sinh(\pi)}{\pi} = \frac{-e^{-\pi}}{2\pi} + \frac{e^\pi}{2\pi}\\
\prod_{n=2}^\infty 1-\frac{1}{n^2} &= \frac{1}{2}
\end{aligned}
$$
A: Assuming you can use the fact that $n! \to \infty$ it follows that $\frac{1}{n!} \to_n 0$ and since $\frac{1}{n!^2} \leq \frac{1}{n!}$ it converges to 0 by squeeze lemma
