1
$\begingroup$

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.

$\endgroup$
11
  • $\begingroup$ $\zeta(s) = P(s)+(\zeta(s)-P(s))$. $\endgroup$
    – reuns
    Commented Dec 17, 2017 at 14:36
  • $\begingroup$ 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/… $\endgroup$
    – Agno
    Commented Dec 17, 2017 at 23:46
  • $\begingroup$ @reuns Thank you for your reply. Could you please elaborate more on your comment? $\endgroup$
    – Klangen
    Commented Dec 18, 2017 at 9:21
  • $\begingroup$ ??? Elaborate on what ? Don't you see how $\sum_{n \text{ composite}} n^{-s}$ appears from what I wrote ? $\endgroup$
    – reuns
    Commented Dec 18, 2017 at 9:23
  • $\begingroup$ @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. $\endgroup$
    – Klangen
    Commented Dec 18, 2017 at 9:33

1 Answer 1

1
$\begingroup$

The identity does not hold (see here). Moreover, the paper that claims this contains a number of (major) problems.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .