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?


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 $$ \lambda(n)=\lambda(p_1^{e_1}\cdots p_r^{e_r})=(-1)^{\sum_{i=1}^re_i}. $$

  • $\begingroup$ ...and f.w.i.w. $= \prod_p (\sum_{n=0}^{\infty }(-1)^n p^{-ns})$ $\endgroup$ – daniel Mar 14 at 12:19

This is just an extended comment to indicate that there is a different way to derive a infinite series sum to replace the product $P_3=\prod _{k=1}^{\infty } \left(\frac{1}{\left(p_k\right){}^3}+1\right)=\frac{\zeta(3)}{\zeta(6)}$ which I found some time ago and hadn't realised the application here.

Starting with $$\sum _{k=1}^{\infty } \frac{\lambda(2 k)}{k (k+1)}=\ln2 + \frac{7 \,\zeta(3)}{2\pi^2}$$

where $\lambda(n)=\sum_{k=1}^\infty\frac{1}{(2k-1)^n}$ is the Dirichlet Lambda Function.

Then using $$\sum _{k=1}^{\infty } \frac{\lambda(2 k)}{2^{2k-1}k}=\ln2$$ to substitute for $\ln(2)$ we eventually arrive at

$$\zeta(3)=\frac{1}{2}\frac{8}{7} \left(\frac{\pi}{2}\right)^3 \left(\frac{\pi}{4}\right)^3 \left( b_{1} \lvert B_{4}\rvert + b_{2} \lvert B_{6} \rvert \left(\frac{\pi}{4}\right)^2 + b_{3} \lvert B_{8} \rvert \left(\frac{\pi}{4}\right)^4+ \; ... \right)$$

where $ b_k=\left( \frac{2^{2k+2}\left( 2^{2k+2}-1\right)\left( 2^{2k+1}-\left( k+2\right)\right)}{\left( 2k+2\right)! \; \left( k+1\right)\left( k+2\right)} \right)$ and where $\lvert B_{2k+2}\rvert =\left(-1 \right)^k \; B_{2k+2}$ are unsigned Bernoulli Numbers.

Which leads to an infinite series for $P_k$ where $\zeta(6)=\frac{\pi^6}{945}$ can be substituted in to give

$$P_3=\frac{945}{896} \left( b_{1} \lvert B_{4}\rvert + b_{2} \lvert B_{6} \rvert \left(\frac{\pi}{4}\right)^2 + b_{3} \lvert B_{8} \rvert \left(\frac{\pi}{4}\right)^4+ \; ... \right)$$

This is a very slowly converging series and you have to be careful when calculating the sum numerically.

I think similar mathematical manoeuvres may be possible with higher odd Zeta(n) constants.


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