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$$\sum_{n=1}^{\infty} \frac{λ(n)}{n^s}=\frac{ζ(2s)}{ζ(s)}$$

Let $λ(n) = (−1)^k$, where $k$ is the number of prime factors of $n$, counting multiplicities. (Liouville function) for $Re(s)>1$, where $ζ(s)$ is the Riemann zeta function.

I am trying to prove this and I don't know where to start.

Hannah
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1 Answers1

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Since $\lambda$ Is multiplicative:

$$\begin{align}\sum_{n=1}^\infty \frac{\lambda(n)}{n^s}&=\prod_p\left(\sum_{k=0}^{\infty} \frac{\lambda(p^k)}{p^{ks}}\right)\\ &=\prod_p \left(\sum_{k=0}^{\infty} \frac{(-1)^k}{p^{ks}}\right)\\ &=\prod_p \left(\frac1{1+\frac 1{p^s}}\right)\\ &=\prod_p \frac{1-\frac1{p^s}}{1-\frac1{p^{2s}}}\\ &=\prod_p \dfrac{\dfrac{1}{1-\frac1{p^{2s}}}}{\dfrac1{1-\frac1{p^s}}}\\ &=\frac{\zeta(2s)}{\zeta(s)} \end{align}$$

Thomas Andrews
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  • As for the first step: symbolically $$ \sum_n n = \prod_p \sum_{k=0} p^k $$ and similarly for and (completely) multiplicative function $f$ : $$ \sum_n f(n) = \prod_p \sum_{k=0} f(p^k) $$ – Raphael J.F. Berger Jul 09 '21 at 14:54
  • It’s the standard product formula for any the Dirichlet function of any multiplicative function in general: $$\sum_{n=1}^\infty \frac{f(n)}{n^s}=\prod_p\left(\sum_{k=0}^{\infty} \frac{f(p^k)}{p^{ks}}\right)$$ if $f$ is a multiplicative function. @RaphaelJ.F.Berger – Thomas Andrews Jul 09 '21 at 14:57
  • @RaphaelJ.F.Berger your formula for $$\sum_n n$$ doesn’t make any sense, or isn’t useful, since none of those series converge. But yes, if $f$ is multiplicative and $\sum f(n)$ converges, your formula is correct. – Thomas Andrews Jul 09 '21 at 15:00
  • Was just a heuristic motivation, thus "symbolically" or to put it in another way assume $f=id$ but the series still converges. I guess there is one or another way to formalize it. – Raphael J.F. Berger Jul 09 '21 at 15:06
  • How do you get from the second to the third line in your procedure? – EGME Aug 04 '22 at 11:33