Given that
$$\zeta(n)=\sum_{k=1}^\infty \frac{1}{k^n}=\prod_{k=1}^\infty \frac{1}{1-\frac{1}{(p_k)^n}}\tag{1}$$
where $n>1$ and $p_k$ is the $k^{th}$ prime.
Proof of the Euler product formula for the Riemann zeta function
It immediately follows that
$$\frac{\zeta(2n)}{\zeta(n)}=\prod_{k=1}^\infty \frac{1}{1+\frac{1}{(p_k)^n}}\tag{2}$$
The question is: Does equation (2) have an easily derivable infinite series form?
Best Answer
Yes, for all $s\in \Bbb C$ with $\Re(s)>1$ we have $$ \frac{\zeta(2s)}{\zeta(s)}=\sum_{n=1}^{\infty}\lambda(n)n^{-s}. $$ Here $\lambda(n)$ is the Liouville function, defined by $\lambda(1)=1$ and $$ \lambda(n)=\lambda(p_1^{e_1}\cdots p_r^{e_r})=(-1)^{\sum_{i=1}^re_i}. $$