Representing $\prod_{k=2}^{\infty}\left(1+\frac{1}{k^{2}-1}\right)$ as an Euler's product.

Question

The above formula corresponds to the following identity

$$ 2=\prod_{k=2}^{\infty}\left(1+\frac{1}{k^{2}-1}\right) $$

I wonder if this can be represented as an Euler's product.

Could anyone find such a representation?

Thanks.

Answer 1

You can expand your product as $$ \begin{align} \prod_{k\ge 2} \left(\frac{1}{1-k^{-2}}\right) &= \prod_{k\ge 2}\left(1+k^{-2}+k^{-4}+k^{-6}+\cdots\right) \\ & = 1 + \frac{1}{2^2}+\frac{1}{3^2}+\frac{2}{4^2}+\frac{1}{5^2}+\frac{2}{6^2}+\cdots \\ & = \sum_{n=1}^\infty \frac{a(n)}{n^2} \end{align} $$

Here $a(n)$ is the number of unordered ways to factor $n$, for example $12^{-2}$ will appear as $2^{-4}3^{-2},2^{-2}6^{-2},3^{-2}4^{-2}$ and $12^{-2}$, so $a(12)=4$.

This $a(n)$ isn't multiplicative, so this expansion does not yield an Euler product in the sense of Wikipedia.

Of course there are products over the primes that equal $2$, e.g. $\prod_p (1+\delta_{2,p})$ where $\delta$ is the Kronecker delta, but I don't think that's what you meant.