# Proving an identity involing the Riemann zeta function

I recently read somewhere (but lost the source) that one could express the composite zeta function $C(n)$ using the prime zeta function $P(n)$ and the Riemann zeta function $\zeta(n)$ in the following manner:

$$C(n)=P(n)\left(\zeta(n)-1\right) - \zeta(n)\sum\frac{1}{p^nq^n},$$

where $n$ is an integer greater than $1$, and the sum $\sum\frac{1}{p^nq^n}$ is taken over all primes $p$ and $q$, for which $p\neq q$. As usual, here $\zeta(n)=\sum_{k=1}^{\infty}\frac{1}{k^n}$.

At first glance this does not even seem to hold (how does the manipulation of infinite series on the right hand side (multiplication of infinite series) factor out all the prime terms?), however I remember seeing a proof somewhere. Unfortunately I forgot to write it down so now I'm trying to work it out myself.

We can expand the left hand side as:

$$\sum_p\frac{1}{p^n}\sum_{k=2}^{\infty}\frac{1}{k^n} - \sum_{k=1}^{\infty}\frac{1}{k^n}\sum\frac{1}{p^nq^n},$$

but I don't see how this results in

$$\sum_c\frac{1}{c^n},$$

i.e., the sum over all composite numbers. Any help is greatly appreciated.

• $\zeta(s) = P(s)+(\zeta(s)-P(s))$. Dec 17, 2017 at 14:36
• On a related matter, maybe of interest. A long while ago I have asked a question on MO about the Composite "Euler" Product: mathoverflow.net/questions/53266/…
– Agno
Dec 17, 2017 at 23:46
• ??? Elaborate on what ? Don't you see how $\sum_{n \text{ composite}} n^{-s}$ appears from what I wrote ? Dec 18, 2017 at 9:23
• @reuns It's obvious that $\zeta(s)=P(s)+C(s)$, that was not my question. Using that in the identity above, however, is not obvious. I would appreciate it if you could elaborate on how it can be used to prove the identity in my question. Dec 18, 2017 at 9:33