How can the sum of reciprocals be derived from the product of primes?

[heartless]

Hello,
I'm trying to prove that

$$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} (1-p^ {-s} )^ {-1}$$

I know why it is and a proof, but I'm actually looking for
a different way to prove going backward and deriving the
sum from the product of primes. Can you show me a way to do that?
I'd like to start with...
$$\prod_{p} (1-p^ {-s} )^ {-1} = ...$$

Thanks,

p.s that -s above and then following -1 should be both exponents for the equation
same with -s and -1 on the bottom, I'm not the master latex writer. Can somebody also tell me why it doesn't work?

 

[SirArthur333]

Hello,
I'm trying to prove that

$$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p} (1-p^ {-s} )^ {-1}$$

I know why it is and a proof, but I'm actually looking for
a different way to prove going backward and deriving the
sum from the product of primes. Can you show me a way to do that?
I'd like to start with...
$$\prod_{p} (1-p^ {-s} )^ {-1} = ...$$

Thanks,

p.s that -s above and then following -1 should be both exponents for the equation
same with -s and -1 on the bottom, I'm not the master latex writer. Can somebody also tell me why it doesn't work?

The golden key lives in the golden house with goldielocks. It is guarded by two dragons. Penitent man will pass.

 

[heartless]

The golden key lives in the golden house with goldielocks. It is guarded by two dragons. Penitent man will pass.

I will never pass... but! I wouldn't have to, if you show me the very proof :)

 

[shmoe]

I know why it is and a proof, but I'm actually looking for
a different way to prove going backward and deriving the
sum from the product of primes. Can you show me a way to do that?

What proof do you know? The usual is to look at the finite product $$\prod_{p\leq x}(1-p^{-s})^{-1}$$ and expand using geometric series. Then compare with $$\sum_{n\leq x} n^{-s}$$, and show the difference the two goes to zero as x-> infinity.