How does one prove the following identity in a combinatorial sense $$ \zeta(s) = \exp \left(\sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} \right)$$

where $\zeta(s)$ is the Riemann zeta function and $\Lambda(n)$ is the Von Mangoldt function. By definition, $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$$

What does $\zeta(s)$ count? Can it be interpreted in terms of a probability? Also can be regard the right hand side as an exponential generating function?

  • $\begingroup$ Don't you want $\zeta$, not $\xi$? $\endgroup$ – Thomas Andrews Jul 22 '13 at 15:45
  • $\begingroup$ You've been reading Terence Tao's recent blog post? $\endgroup$ – Andreas Caranti Jul 22 '13 at 16:31
  • $\begingroup$ @AndreasCaranti: yes $\endgroup$ – guestgoijeoiejr Jul 22 '13 at 16:33

I have an idea for a proof, though I don't know if this qualifies to be a combinatorial proof at all. Still I'm giving this proof. Pardon me if it is not even close to what is desired.

Since $$\Lambda(n)=\left\{\begin{array}{rl} \log p & \mbox{if}\ n=p^k\\ 0 & \mbox{else} \end{array} \right.$$ We can rewrite the RHS as the sum over all numbers of the form $p^k$ where $p$ are primes. \begin{equation} \begin{split} \sum_{n\ge 1}\frac{\Lambda(n)}{\log n}n^{-s}=& \sum_{k\ge 1}\sum_{p}\frac{\Lambda(p^k)}{\log(p^k)}(p^k)^{-s}\\ \ =& \sum_{k\ge 1}\sum_{p}\frac{\log p}{k\log p}p^{-ks}\\ \ =& \sum_{k\ge 1}\sum_{p}\frac{1}{k}p^{-ks}\\ \end{split} \end{equation}

Now, we can observe that, by Euler product \begin{equation} \begin{split} \zeta(s)=&\frac{1}{\displaystyle \prod_{p}\left(1-p^{-s}\right)}\\ \Rightarrow \log(\zeta(s))=& -\sum_{p}\log(1-p^{-s})\\ \ =&\sum_{p}\sum_{k\ge 1}\frac{p^{-ks}}{k} \end{split} \end{equation}

Hence $$\log(\zeta (s))=\sum\frac{\Lambda(n)}{\log n}n^{-s}\\ \Rightarrow \zeta(s)=\exp\left(\sum\frac{\Lambda(n)}{\log n}n^{-s}\right) \Box $$


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