$$\frac{\zeta(s)}{\zeta(hs)} =\prod_p\left(\frac{1-\frac{1}{p^{hs}}}{1-\frac{1}{p^{s}}}\right) =\prod_p\left(1+\frac{1}{p^s}+\cdots +\frac{1}{p^{(h-1)s}}\right)=\sum_{n\in S_h}\frac{1}{n^s}$$
What is the step by step process proof of this series identity?
Hs is for h-free numbers, where in by definition it is positive integer n is square-free if and only if its prime factorization has no factors with an exponent larger than one
I am stuck here:
$$\frac{\zeta(s)}{\zeta(hs)}=\prod_p\frac{\left(\frac{1}{1-\frac{1}{p^s}}\right)}{\left(\frac{1}{1-\frac{1}{p^{hs}}}\right)} =\prod_p\left(\frac{1-\frac{1}{p^{hs}}}{1-\frac{1}{p^{s}}}\right)$$
Would it be possible to come up with this?
$$\frac{1-\frac{1}{p^{hs}}}{1-\frac{1}{p^s}}=\frac{\left(\frac{p^{hs}-1}{p^{hs}}\right)}{\left(\frac{p^s-1}{p^s}\right)}=\frac{p^s(p^{hs}-1)}{p^{hs}(p^s-1)}$$
Correct me if I'm wrong.
my question is that, how it leads to reimann zeta function?
You probably need this formula:
For every $x\in\mathbb{R}$,
$$1-x^n = (1-x)(1+x+...+x^{n-1})$$
Which you can prove by induction. It follows that for all $x\not = 1$, $$ \frac{1-x^n}{1-x} = 1+x+...+x^{n-1}$$ Now just plug in $x=\frac{1}{p^s}$.