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?


  • $\begingroup$ @ShreevatsaR, I know this, but what I'm asking is if this can also be expressed as an infinit product over prime numbers. $\endgroup$ – Neves Apr 23 '13 at 16:08
  • $\begingroup$ @joriki: Oops. Of course. :) Actually, in light of the OP's comment, I've deleted both my comments. $\endgroup$ – ShreevatsaR Apr 23 '13 at 16:52
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    $\begingroup$ I do not understand the question. Euler product is one which converts a series, i.e., a summation into a product of a similar summation over primes. Here you have a product, so could you be clear what you expect? $\endgroup$ – user17762 Apr 23 '13 at 17:10
  • $\begingroup$ @user17762, suppose that this product is expanded into an infinite sum. My question is: Can we get an infinite product over primes, a kind of Euler's product like we have for $\pi^2/6$? $\endgroup$ – Neves Apr 23 '13 at 17:14
  • $\begingroup$ I mean $\zeta(2)=\pi^2/6=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^2-1}\right)$. $\endgroup$ – Neves Apr 23 '13 at 17:22

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.


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