proof with Euler products

Asked May 25 '20 at 13:58

Active May 25 '20 at 13:58

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Let $\zeta(s)$ be the Riemann zeta function.$$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.$$ Then we know it satisfies the Euler product for $\text{Re}(s) > 1$,

$$\zeta(s) = \prod_{p} (1 - p^{-s})^{-1}.$$ I need to use Euler products to proof that

$$\frac{\zeta(2s)}{\zeta(s)} = \prod_{p} (1 + p^{-s})^{-1}. $$

I know that $$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{μ(n)}{n^s} $$ (μ is mobius function)

I am really don't know what to do from here.

asked May 25 '20 at 13:58

qqq aaa

Use the Euler product given here. – Dietrich Burde May 25 '20 at 14:06
1

$Is 1-p^{-2s}=(1-p^{-s})(1+p^{-s})$? – N.Quy May 25 '20 at 14:07
thank you very much. it's so simple that it's genius – qqq aaa May 25 '20 at 14:17
qqqaaa Would be good if you use @N.Quy hint to write a (two lines) answer. – reuns May 25 '20 at 14:58

proof with Euler products

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