# Where does this series converge $\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}$, being $\mu(n)$ the Möbius function?

Let $\mu(n)$ the Möbius function and $s=\sigma+it$ the complex variable, then I've defined the Dirichlet series $$\epsilon(s):=\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}.$$ And now I know that using absolute convergence it converges for $\Re s>1$, since $$\left|\sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}\right|\leq\sum_{n=1}^\infty\frac{1}{n^{\Re s}}.$$

Question. Can you improve this abscissa where the series is convergent? If it is not feasible explain why, if your explanation is possible. Thanks in advance.

• $(-1)^{\mu(n)}$ is $-1$ if $n=1$ or $n$ is square-free, $1$ if $n$ is not square-free. – Sungjin Kim Mar 14 '17 at 1:27
• @i707107 many thanks for your attention, I believe that you are saying that there are Dirichlet series corresponding these definitions that could help us. My Dirichlet series was an invention, and now I don't know if are feasibles your calculations. – user243301 Mar 14 '17 at 1:39

That is not a Dirichlet series strictly speaking, since $(-1)^{\mu(n)}$ is not a multiplicative function.
We have $$(-1)^{\mu(n)}=1-2\mu(n)^2$$ and square-free numbers have a positive density among integers, hence $$\sum_{n\geq 1}\frac{(-1)^{\mu(n)}}{n^s} = \zeta(s)-2\sum_{n\geq 1}\frac{\mu(n)^2}{n^s} = \zeta(s)-\frac{2\,\zeta(s)}{\zeta(2s)}$$ and the abscissa of convergence is one.

• I hope I'm not bored, but can you explain more the reasoning from last identity, the why it is obvious that then the abscissa of convergence is $1$. That is you are taking a limit or seeing the domain of definiton of the functions? – user243301 Mar 14 '17 at 1:49
• @user243301: Such a series can be computed fairly easy for any $s>1$ and the Riemann $\zeta$ function has a simple pole at $s=1$. – Jack D'Aurizio Mar 14 '17 at 1:51
• Yes, I understand now, many thanks. – user243301 Mar 14 '17 at 1:52

The density of squarefree integers is $6/\pi^2$, so about 60% of the integers up to $X$ are squarefree (and this is very accurate for large $X$). So 60% of the time, your summands are $-1/n^s$, while 40% of the time your summands are $1/n^s$.

At $s = 1$, the expected value of your sum up to $X$ is $-\frac{1}{5}X$, which clearly diverges. Therefore the abscissa of convergence is $1$.

• Many thanks seems a magic argument, I am saying it seriously that with an heuristic argument you can prove the abscissa of convergence. Many thanks, is great. – user243301 Mar 14 '17 at 1:46
• I get that $$\sum_{n\leq X}\frac{(-1)^{\mu(n)}}{n}\approx \left(1-\frac{12}{\pi^2}\right)\log X$$ – Jack D'Aurizio Mar 14 '17 at 1:47
• I suppose I could write my actual heuristic more precisely. As $6X/\pi^2$ summands up to $X$ give $-1$ and $(1-6/\pi^2)X$ give $1$, the expected value should be $-6X/\pi^2 + (1 - 6/\pi^2)X = (1 - 12/\pi^2)X$. Oh - I was computing at $s = 0$. Whoops. Now the heuristic is a bit more demanding intuitively. But if one pretends that $1/n$ is about the same size for many $n$ near some large $N$, then one might expect $(1 - 12/\pi^2)$ percent of the terms to contribute from my heuristic, giving exactly $(1 - 12/\pi^2)\log X$. Making this rigorous is morally doable, but more annoying than your answer. – davidlowryduda Mar 14 '17 at 1:58
• This is a present for you. Is a simple application of your answer and this MSE. Then I believe that I can write $\lim_{n\to\infty}\sum_{k=1}^n (-1)^{\mu(k)}/n^{1+\epsilon}=0$ for each fixed $\epsilon>0$. A direct proof from Jack's $(-1)^{\mu(n)}=1-2\mu(n)^2$ and a well known asymptotic for square-free tell us that the limit is $\lim_{n\to\infty}n^{-\epsilon}-2\left(\frac{1}{\zeta(2)}n^{-\epsilon}+O\left(n^{-1/2-\epsilon}\right)\right)=0$. – user243301 Mar 14 '17 at 22:28
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