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Recently I've been learning about the basics of Dirichlet functions and their relations to subjects in number theory that I would characterise as combinatorial in nature. I came across this identity: $$\frac{\zeta(s)^2}{\zeta(2s)}=\sum_{n\geq1}2^{\nu(n)}n^{-s},$$ Here $\zeta$ denotes the Riemann zeta function, and $\nu(n)$ is the number of distinct prime divisors of $n$. I'm able to prove that this identity holds (for all $s\in\Bbb{C}$ with $\Re (s)>1$) by writing both sides as Euler products and rearranging terms. However, I also noticed that $$\zeta(s)^2=\sum_{n\geq1}\sigma_0(n)n^{-s}\qquad\text{ and }\qquad\sum_{n\geq1}2^{\nu(n)}n^{-s}=\sum_{n\geq1}\sigma_0(\operatorname{rad}(n))n^{-s},$$ where $\sigma_0(n)$ denotes the number of positive divisors of $n$ and $\operatorname{rad}(n)$ denotes the radical of $n$. This suggests to me that there might be some sort of counting argument for the identity above.

Unfortunately I'm unable to make such an argument work. I've tried rewriting the left hand side in a few ways, for example using $$\zeta(2s)^{-1}=\sum_{n\geq1}\mu(n)n^{-2s}=\sum_{n\geq1}\tau(n)n^{-s},$$ where $\mu$ denotes the Möbius function and $\tau$ is rather ugly, see below, to get $$\frac{\zeta(s)^2}{\zeta(2s)}=\sum_{n\geq1}\left(\sum_{d\mid n}\tau(d)\sigma_0(\tfrac{n}{d})\right)n^{-s},$$ so I'd like to show that $\sum_{d\mid n}\tau(d)\sigma_0(\tfrac{n}{d})=\sigma_0(\operatorname{rad}(n))$, where $\tau$ is given by \begin{eqnarray*} \tau(n) &:=& \left\{\begin{array}{ll} \mu(\sqrt{n})&\text{ if $n$ is a square}\\ 0&\text{ if $n$ is not a square} \end{array}\right.\\ &=& \left\{\begin{array}{ll} \mu(\operatorname{rad}(n))&\text{ if }n=\operatorname{rad}(n)^2\\ 0&\text{ otherwise }\end{array}\right.\\ &=& \left\{\begin{array}{ll} (-1)^{{}^2\log\sigma_0(\operatorname{rad}(n))}&\text{ if }n=\operatorname{rad}(n)^2\\ 0&\text{ otherwise }\end{array}\right.\\ \end{eqnarray*} But I can't quite seem to fit these ingredients together. A hint as to how to write $\tau$ and/or how to rewrite the sum $\sum_{d\mid n}\tau(d)\sigma_0(\tfrac{n}{d})$ would be much appreciated. This is homework, by the way.

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Start from $$2^{\nu(n)} = \sigma_0(rad(n)) = \sum_{d | rad(n)} 1 = \sum_{d | n} |\mu(d)|$$ and use $$|\mu(n)| = \sum_{d^2 | n} \mu(d)$$

so that $$2^{\nu(n)} = \sum_{d | n} \sum_{k^2 | d} \mu(k) = \sum_{k^2 | n} \mu(k)\sum_{d | n} 1_{k^2 | d} = \sum_{k^2 | n} \mu(k)\sigma_0(n/k^2)$$

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  • $\begingroup$ @Servaes using Euler products yes :) $\endgroup$ – reuns Oct 24 '16 at 15:34
  • $\begingroup$ @Servaes If you know that $\sum_{d | n} \mu(d) = 1_{n =1}$ then let $L(n) = \max_{m, m^2 | n} m^2$ and you get $\sum_{d^2 | n} \mu(d) = \sum_{d | L(n)} \mu(d) = 1_{L(n) = 1}$ which is $|\mu(n)|$ if you know that $\mu(n) = (-1)^{\nu(n)} 1_{L(n) = 1}$ $\endgroup$ – reuns Oct 24 '16 at 16:22
  • $\begingroup$ So to be clear you need the 2 definitions of $\mu$ : as the Dirichlet inverse of $1$, and as $(-1)^{\nu(n)} 1_{L(n) = 1}$ $\endgroup$ – reuns Oct 24 '16 at 16:23
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Since $$ \zeta(s)=\prod_{p\in\mathbb{P}}\frac1{1-\frac1{p^s}} $$ we also have $$ \frac{\zeta(s)}{\zeta(2s)}=\prod_{p\in\mathbb{P}}\left(1+\frac1{p^s}\right) $$ Therefore, $$ \begin{align} \frac{\zeta(s)^2}{\zeta(2s)} &=\prod_{p\in\mathbb{P}}\frac{p^s+1}{p^s-1}\\ &=\prod_{p\in\mathbb{P}}\left(1+\frac2{p^s-1}\right)\\ &=\prod_{p\in\mathbb{P}}\left(1+\frac2{p^s}+\frac2{p^{2s}}+\frac2{p^{3s}}+\cdots\right)\\ &=\sum_{n=1}^\infty2^{\nu(n)}n^{-s} \end{align} $$

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    $\begingroup$ Sorry. I noticed that part after I wrote my answer. However, it seemed to be a useful comparison to the answer youi're looking for, so I left it. $\endgroup$ – robjohn Oct 24 '16 at 10:31

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