Does the Euler product stand for $a(n)=rad(n)$? Or more generally, for multiplicative functions which are not completely multiplicative?

Where rad is the product of a number's distinct prime factors.

Wikipedia states that we can calculate the Euler product where $a(n)$ is multiplicative but doesn't stipulate completely multiplicative.

Rad function is multiplicative but not completely multiplicative.

So we would have:

$\displaystyle\sum_{n=0}^{\infty}rad(n)n^{-s}=\prod_p \left(1+rad(p)p^{-s}+rad(p^2)p^{-2s}+rad(p^3)p^{-3s}+\ldots\right)\\=\prod_p 1+p^{1-s}+p^{1-2s}\ldots=\prod_p \left(p(1+p^{1-s}+p^{1-2s}\ldots)+1-p\right)$

... there's a bit more work needed to relate this back to $\zeta(s)$ but I want to be sure the Euler product is valid first.

  • 1
    $\begingroup$ Sure iff $a(n)$ is multiplicative then $\sum_{n=1}^\infty a(n) n^{-s} = \prod_p (1+\sum_{k \ge 1} a(p^k) p^{-sk})$. If the LHS converges for $\Re(s)$ large enough then so does the RHS. With $ rad(n)$ you get $$\sum_{n=1}^\infty rad(n) n^{-s} = \prod_p (1+\frac{p^{1-s}}{1-p^{-s}})= \zeta(s) \prod_p (1+(p-1)p^{-s}))=\zeta(s) \sum_{n=1}^\infty |\mu(n)| \phi(n)n^{-s}$$ which converges for $\Re(s) > 2$ and allows to estimate things like $\sum_{n \le x} rad(n) n^{-s} \log^k n$ using the same method as for the PNT $\endgroup$ – reuns Oct 14 '17 at 1:46

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