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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.

The motivation of this series is to learn more about convergence and calculations with series.

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    $\begingroup$ $(-1)^{\mu(n)}$ is $-1$ if $n=1$ or $n$ is square-free, $1$ if $n$ is not square-free. $\endgroup$ – Sungjin Kim Mar 14 '17 at 1:27
  • $\begingroup$ @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. $\endgroup$ – user243301 Mar 14 '17 at 1:39
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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.

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  • $\begingroup$ 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? $\endgroup$ – user243301 Mar 14 '17 at 1:49
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    $\begingroup$ @user243301: Such a series can be computed fairly easy for any $s>1$ and the Riemann $\zeta$ function has a simple pole at $s=1$. $\endgroup$ – Jack D'Aurizio Mar 14 '17 at 1:51
  • $\begingroup$ Yes, I understand now, many thanks. $\endgroup$ – user243301 Mar 14 '17 at 1:52
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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 $1$.

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  • $\begingroup$ 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. $\endgroup$ – user243301 Mar 14 '17 at 1:46
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    $\begingroup$ I get that $$\sum_{n\leq X}\frac{(-1)^{\mu(n)}}{n}\approx \left(1-\frac{12}{\pi^2}\right)\log X$$ $\endgroup$ – Jack D'Aurizio Mar 14 '17 at 1:47
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    $\begingroup$ 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. $\endgroup$ – davidlowryduda Mar 14 '17 at 1:58
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    $\begingroup$ 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 $\lim_{n\to\infty}n^{-\epsilon}-2\left(\frac{1}{\zeta(2)}n^{-\epsilon}+O\left(n^{-1/2-\epsilon}\right)\right)=0$. $\endgroup$ – user243301 Mar 14 '17 at 22:28

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