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This is going to be an alternative answer to Dr. Wolfgang Hintze's question. Consider a limit: $$g := \prod\limits_{n=1}^\infty \frac{1}{\left(1-\frac{A^2}{n^2}\right)\left(1-\frac{B^2}{n^2}\right)\left(1-\frac{C^2}{n^2}\right)\left(1-\frac{D^2}{n^2}\right)}$$ Taking logs we have: \log(g) = - ...

1

This is not an answer. I have additional observations about the answer by Dr. Wolfgang Hintze. In response to his first comment: From here, when we omit the first two primes from Euler's product, we get a square: $$\frac{\pi^2}{9} =\prod _{n=3}^{\infty } \frac{1}{1-p_n^{-2}},$$ then substituting $6n$ for $p_n,$ we get the square root: $$\frac{\pi}{3} =\prod ... 7 I tried to find an even simpler product. Here's my solution:$$ \zeta(2) =\prod _{n=1}^{\infty } \frac{1}{\left(1-\frac{1}{4 n^2}\right) \left(1-\frac{1}{36 n^2}\right)} In Mathematica Product[ 1/(1 - 1/(4 n^2)) 1/(1 - 1/(36 n^2)), {n, 1, \[Infinity]}] (* Out[76]= \[Pi]^2/6 *) We can derive this from the well-known product formula of the sine ...

12

Amplifying on answer by @rhermans f[m_] = Product[(1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3)), {n, 1, m}] (* (Pi^2*Gamma[1 + m]^3*Gamma[3 + m])/ (6*(3 + 3*m + m^2)*Gamma[5/6 + m]^ 2*Gamma[7/6 + m]^2) *) This product converges Limit[f[m + 1]/f[m], m -> Infinity] (* 1 *) Limit[f[m], m -> Infinity] (* Pi^2/6 ...

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Product[ (1296 n^4 (1 + (1 + n)^3))/((-1 + 36 n^2)^2 (-1 + (1 + n)^3)) , {n, 1, ∞} ] === Zeta[2] True Zeta[2] π^2/6

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