In Wolfram MathWorld site at Moebius Function topic there is identity number 10, which states that $\sum_{n=1}^\infty \frac{\mu(n)}{n} = 0$. Could you help me find a rigorous proof of this statement in the literature? Any help would be appreciated!
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12$\begingroup$ Landau's PhD thesis: arxiv.org/abs/0803.3787 $\endgroup$– Micah MilinovichCommented Oct 11, 2021 at 16:45
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1$\begingroup$ @MicahMilinovich, I think that's an answer! $\endgroup$– LSpiceCommented Oct 11, 2021 at 16:45
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2$\begingroup$ math.stackexchange.com/q/1169675/127263 $\endgroup$– WojowuCommented Oct 11, 2021 at 17:17
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3$\begingroup$ @JosephVanName This argument (together with an appropriate version of Abel's theorem for Dirichlet series) shows that if the limit exists, then it is equal to $0$. However, convergence is far from clear. Questions like this are subtle and convergence depends on distribution of zeros of zeta. For instance, convergence for all $s>1/2$ is equivalent to RH. $\endgroup$– WojowuCommented Oct 11, 2021 at 17:46
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2$\begingroup$ You are right. I was not being rigorous. $\endgroup$– Joseph Van NameCommented Oct 11, 2021 at 17:57
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1 Answer
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Ayoub, An Introduction to the Analytic Theory of Numbers: Theorem 7.10 on p. 111.
Bateman and Diamond, Analytic Number Theory: An Introductory Course: Theorem 5.9 on p. 100.
Hlawka, Taschner, and Schoissengeier, Geometric and Analytic Number Theory: Corollary 4 on p. 123.
Jameson, The Prime Number Theorem: Theorem 3.4.5 on p. 134.
Montgomery and Vaughan, Multiplicative Number Theory I: Classical Theory: equation (8.8) on p. 248.
Narkiewicz, The Development of Prime Number Theorey from Euclid to Hardy and Littlewood: Theorem 5.20 on p. 239.
Overholt, A Course in Analytic Number Theory: pp. 52-54 shows convergence of $\sum \mu(n)/n$ implies the sum is $0$, the convergence of $\sum \mu(n)/n$ is equivalent to $(1/x)\sum_{n \leq x} \mu(n) \to 0$, and that limit being $0$ is equivalent to the Prime Number Theorem.
Titchmarsh, The Theory of the Riemann Zeta-Function (2nd ed.): Theorem 3.13 on p. 62.
In the third and fourth references, the proof is essentially based on the following theorem. If the Dirichlet series $f(s) = \sum_{n \geq 1} b_n/n^s$ has bounded coefficients, so it converges for ${\rm Re}(s) > 1$, and $f$ has an analytic continuation to the line ${\rm Re}(s) = 1$ except for at worst a simple pole at $1$, with $$ f(s) = \frac{\rho}{s-1} + c + O(s-1) $$ near $s = 1$, then $(1/x)\sum_{n \leq x} b_n \to \rho$ and $\sum_{n \leq x} b_n/n = \rho\log x + c + o(1)$ as $x \to \infty$. In particular, if $\rho = 0$, meaning $f$ is analytic at $s = 1$ and $c = f(1)$, then $(1/x)\sum_{n \leq x} b_n \to 0$ as $x \to \infty$ and $f(1) = \sum_{n \geq 1} b_n/n$. Taking $f(s) = \sum \mu(n)/n^s = 1/\zeta(s)$, we get $(1/x)\sum_{n \leq x} \mu(n) \to 0$ and $\sum_{n \geq 1} \mu(n)/n = 0$ since $1/\zeta(s)$ is holomorphic on ${\rm Re}(s) \geq 1$ with a zero at $s = 1$.
