# Zeta(3) and Euler's formula

Hi everyone. I'm trying to understand the step where they wrote

1/2 ∏1/(1+p^-3) =1/2 Ʃ(-1)^ord(k)/k^3

How can I see this? I know the Euler product formula, but it has a negative sign before the p^-3, where here we have a + sign.

Thanks for the help.

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The step going from the second to last line, to the last line.

I need help understanding this equality:

$\prod_{p-prime} \frac{1}{1+\frac{1}{p^3}}= \sum_{k=1}^\infty \frac{(-1)^{\sum_p ord_p(k)}}{k^3}$

Any help is greatly appreciated!!!

lurflurf
Homework Helper
We know
$$\prod_\mathbb{P} (1-p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} n^{-s}$$

with the minus sign it is a little more complicated

$$\prod_\mathbb{P} (1+p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} (-p)^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} (-1)^{\sum ord_p(k)}n^{-s}$$

Where the ord_p(k) makes sure we get the right sign (it counts the minuses), clearly

$$\left( \prod_\mathbb{P} (1-p^{-s})^{-1} \right) \left( \prod_\mathbb{P} (1+p^{-s})^{-1} \right) =\zeta(2n)$$

We know
$$\prod_\mathbb{P} (1-p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} n^{-s}$$

with the minus sign it is a little more complicated

$$\prod_\mathbb{P} (1+p^{-s})^{-1}=\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} (-p)^{-s n} \right) ^{-1} = \sum_{\mathbb{Z}>0} (-1)^{\sum ord_p(k)}n^{-s}$$

Where the ord_p(k) makes sure we get the right sign (it counts the minuses), clearly

$$\left( \prod_\mathbb{P} (1-p^{-s})^{-1} \right) \left( \prod_\mathbb{P} (1+p^{-s})^{-1} \right) =\zeta(2n)$$

Should $\prod_\mathbb{P} \left( \sum_{\mathbb{Z} \ge 0} p^{-s n} \right) ^{-1}$ have that ^(-1) after it? Or am I missing something..? Are you rewriting 1/(1-p^-s) using geometric series?

Anyways, thanks that was very helpful, I'm looking into the proofs of the product formula via this route.

lurflurf
Homework Helper
Yes that is a geometric series expansion.

lurflurf
Homework Helper
$$\prod_\mathbb{P} (1+p^{-s})^{-1}= \prod_\mathbb{P} (1-(-p^{-s}))^{-1}= \\ \prod_\mathbb{P} \left( 1-p^{-s}+p^{-2s}-p^{-3s}+...+(-1)^k p^{-ks}+... \right) ^{-1} = \sum_{\mathbb{Z}>0} a_n n^{-s}=\frac{\zeta(2n)}{\zeta(n)}$$
a_n being in{-1,1}
In the sum each term is positive or negative and we can determine which using
ord_p(k)
it is a bit tedious to compute, which is why the link estimates
here is the wikipedia
[PLAIN]http://en.wikipedia.org/wiki... one way (fundamental theorem of arithmetic).

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