here is the following theorem:

"Given the two functions F(s) and G(s) represented by Dirichlet series with $L(s,f)=\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$ for $\sigma>a$ and $L(s,g)=\sum_{n=1}^{\infty} \frac{g(n)}{n^s}$ for $\sigma>b$
then in the half plane where both series converge absolutely we have

L(s,h)=L(s,f)L(s,g)=$\sum_{n=1}^{\infty} \frac{h(n)}{n^s}$ with h=f*g. (convolution of f and g)"

Here is my issue: We know that L(s,$\mu$)=1/$\zeta(s)$ and L(s,$\mu^2$)=$\zeta(s)/\zeta(2s)$ converge absolutely for Re(s)>1. Suppose I want to find L(s, $\mu * \mu^2$) for s=1. We can see that L(s,$\mu$)L(s,$\mu^2$)=1/$\zeta(2s)$ and one can easily see that 1/$\zeta(2s)$ converges absolutely for Re(s)>1/2. (so their product converges absolutely at s=1, even if L(s,$\mu$) and L(s,$\mu^2$) do not.)

Can I still apply the theorem to conclude that L(1, $\mu * \mu^2$) = 1/$\zeta(2*1)$=1/$\zeta(2)$?


  • $\begingroup$ I don't get your question. with $a = \mu \ast |\mu| $ you have that $\sum_{n=1}^\infty a_n n^{-s} = \frac{\zeta(s)}{\zeta(2s)}\frac{1}{\zeta(s)} = \frac{1}{\zeta(2s)} = \sum_{n=1}^\infty \mu_n n^{-2s}$ i.e. $a_{n^2} = \mu_n, \ a_n = 0$ otherwise, and it converges absolutely for $Re(s)> 1/2$ $\endgroup$ – reuns Jul 13 '16 at 1:38
  • 1
    $\begingroup$ maybe your confusion comes from $\lim_{N \to \infty} \sum_{n=1}^N a_n n^{-s} = \frac{1}{\zeta(2s)}$ for $Re(s)> 1/2$, but $\lim_{N \to \infty} (\sum_{n=1}^N \mu_n n^{-s})(\sum_{n=1}^N |\mu_n| n^{-s})$ diverges for $Re(s) < 1$. $\ \ $ this is because $\sum_{n=1}^N (u \ast v)_n n^{-s} = \sum_{n=1}^N u_n n^{-s}\sum_{k=1}^{N/n} v_k k^{-s} \ne (\sum_{n=1}^N u_n n^{-s})(\sum_{n=1}^N v_n n^{-s})$ $\endgroup$ – reuns Jul 13 '16 at 1:42
  • $\begingroup$ My question is that, in order to use the theorem, i.e. use the fact that L(1,$\mu*\mu^2$)=L(1,$\mu$) L(1,$\mu^2$), it is necessary that L(1,$\mu$) and L(1,$\mu^2$) both converge absolutely at s=1. But they only converge for Re(s)>1. $\endgroup$ – usere5225321 Jul 13 '16 at 1:47
  • $\begingroup$ no, read what I wrote. and if you want, I used that this is true for $Re(s) $ large enough (for proving that $a_{n^2} = \mu(n)$). but I could also use that $a_n$ is multiplicative, and show that $a_{p^{2k}} = \mu \ast |\mu|(p^{2k}) = \sum_{d | p^{2k}} \mu(d) |\mu(p^{2k}/d)| = \mu(p^{k})$) $\endgroup$ – reuns Jul 13 '16 at 1:49
  • $\begingroup$ and when we write $\frac{\zeta(s)}{\zeta(2s)} \frac{1}{\zeta(s)}=\frac{1}{\zeta(2s)}$ this is true as an equality of en.wikipedia.org/wiki/Meromorphic_function (on the whole complex plane except at the poles) no need their representation as Dirichlet series on a half-plane to converge $\endgroup$ – reuns Jul 13 '16 at 1:54

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