# abscissa of absolute convergence for ∑∞n=1μ2(n)ns\sum_{n=1}^\infty \frac{\mu^2(n)}{n^s}

Define F(s)=∞∑n=1μ2(n)nsF(s) = \sum_{n=1}^\infty \frac{\mu^2(n)}{n^s} where s∈C.s \in \mathbb{C}.

Verify that “If kk is a constant and g:N→Cg : \mathbb{N} \rightarrow \mathbb{C} with g(n)=Oδ(nδ+k)g(n) = O_\delta(n^{\delta + k}) for any δ>0\delta > 0, then the abscissa of absolute convergence, σ2(G)\sigma_2(G), for G(s)=∞∑n=1g(n)nsG(s) = \sum_{n=1}^\infty \frac{g(n)}{n^s} satisfies σ2(G)≤k+1.\sigma_2(G) \leq k+1.”

Then use the statement to show that σ1(F)=σ2(F)=1.\sigma_1(F) = \sigma_2(F) =1.

Note : Let H(s)H(s) denote the Dirichlet series, then σ1(H)=inf{σ∈R:H(σ) converges},\sigma_1(H) = \inf\{\sigma \in \mathbb{R} : H(\sigma) \ \mbox{converges}\}, \sigma_2(H) = \inf\{\sigma \in \mathbb{R} : H(\sigma) \ \mbox{converges absolutely}\}.\sigma_2(H) = \inf\{\sigma \in \mathbb{R} : H(\sigma) \ \mbox{converges absolutely}\}.

\textbf{Proof}\textbf{Proof} Fix \delta > 0\delta > 0, then there exists n(\delta)n(\delta) and a constant c(\delta)c(\delta) such that |g(n)| \leq c(\delta)n^{k + \delta}|g(n)| \leq c(\delta)n^{k + \delta} for any n \geq n(\delta).n \geq n(\delta). So \sum_{n=n(\delta)}^\infty \frac{|g(n)|}{n^\sigma} \leq c(\delta) \sum_{n \geq n(\delta)}\frac{1}{n^{\sigma – k – \delta}}.\sum_{n=n(\delta)}^\infty \frac{|g(n)|}{n^\sigma} \leq c(\delta) \sum_{n \geq n(\delta)}\frac{1}{n^{\sigma – k – \delta}}. The last summation will be convergent if and only if \sigma > k + \delta + 1.\sigma > k + \delta + 1. By definition, \sigma_2(G) \leq k + \delta + 1\sigma_2(G) \leq k + \delta + 1 for any \delta > 0\delta > 0.

Therefore \sigma_2(G) \leq k+1\sigma_2(G) \leq k+1

\textbf{Is my proof correct ?}\textbf{Is my proof correct ?}

\textbf{Proof}\textbf{Proof} Since \frac{\mu^2(n)}{n^\sigma} \geq 0\frac{\mu^2(n)}{n^\sigma} \geq 0 for any \sigma \in \mathbb{R}\sigma \in \mathbb{R}, \sigma_1(F) = \sigma_2(F).\sigma_1(F) = \sigma_2(F).
Since n^xn^x is an incresing function and n^x \rightarrow 1n^x \rightarrow 1 as x \rightarrow 0x \rightarrow 0, \mu^2(n) = O_\delta(n^\delta)\mu^2(n) = O_\delta(n^\delta) for any \delta > 0.\delta > 0. Then \sigma_2(F) \leq 1\sigma_2(F) \leq 1. Since \sum_{p \ \mbox{prime}}\frac{1}{p} \leq F(1)\sum_{p \ \mbox{prime}}\frac{1}{p} \leq F(1) and \sum_{p \ \mbox{prime}}\frac{1}{p}\sum_{p \ \mbox{prime}}\frac{1}{p} diverges, \sigma_2(F) \geq 1.\sigma_2(F) \geq 1.

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Yes. You can also say that since |\mu(n)||\mu(n)| is multiplicative : we have the formal equality \sum_{n=1}^\infty |\mu(n)|n^{-s} = \prod_p (1+\sum_{k=1}^\infty |\mu(p^k)|p^{-sk}) = \prod_p (1+p^{-s})\sum_{n=1}^\infty |\mu(n)|n^{-s} = \prod_p (1+\sum_{k=1}^\infty |\mu(p^k)|p^{-sk}) = \prod_p (1+p^{-s}). Now for s \in \mathbb{R}s \in \mathbb{R}, those are series/products of non-negative terms, so the LHS converges iff the RHS converges, and thanks to \zeta(s) = \prod_p \frac{1}{1-p^{-s}} = \prod_p (1+p^{-s}+\mathcal{O}(p^{-2s}))\zeta(s) = \prod_p \frac{1}{1-p^{-s}} = \prod_p (1+p^{-s}+\mathcal{O}(p^{-2s})) that converges for Re(s) > 1Re(s) > 1 and diverges for Re(s) < 1Re(s) < 1, it means \prod_p (1+p^{-s}) = \sum_{n=1}^\infty |\mu(n)|n^{-s}\prod_p (1+p^{-s}) = \sum_{n=1}^\infty |\mu(n)|n^{-s} converges for Re(s) > 1Re(s) > 1 and diverges for Re(s) < 1Re(s) < 1 – user1952009 Oct 20 at 19:47      \mu^2(n)\mu^2(n) is the characteristic function of squarefree integers, that have a positive density (\frac{6}{\pi^2}\frac{6}{\pi^2}). It follows that the abscissa of convergence of \sum_{n\geq 1}\frac{\mu^2(n)}{n^s}\sum_{n\geq 1}\frac{\mu^2(n)}{n^s} is the same as the abscissa of convergence of \sum_{n\geq 1}\frac{1}{n^s}.\sum_{n\geq 1}\frac{1}{n^s}. – Jack D'Aurizio Oct 20 at 21:02      You may also notice that the associated Euler products are very similar: \prod_p\left(1+\frac{1}{p^s}\right),\qquad \prod_p\left(1-\frac{1}{p^s}\right)^{-1}\prod_p\left(1+\frac{1}{p^s}\right),\qquad \prod_p\left(1-\frac{1}{p^s}\right)^{-1} – Jack D'Aurizio Oct 20 at 21:04      In particular, \sum_{n\geq 1}\frac{\mu^2(n)}{n^s}=\frac{\zeta(s)}{\zeta(2s)}.\sum_{n\geq 1}\frac{\mu^2(n)}{n^s}=\frac{\zeta(s)}{\zeta(2s)}. – Jack D'Aurizio Oct 20 at 21:05 ================= =================