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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.

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 $s = 1$.

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 $s = 1$.

Corollary 4 (p. 123) in Hlawka, Taschner, and Schoissengeier, Geometric and Analytic Number Theory.

Theorem 7.10 (p. 111) in Ayoub, An Introduction to the Analytic Theory of Numbers.

Theorem 3.4.5 (p. 134) in Jameson, The Prime Number Theorem.

Theorem 5.20 (p. 239) in Narkiewicz, The Development of Prime Number Theorey from Euclid to Hardy and Littlewood.

Theorem 5.9 (p. 100) in Bateman and Diamond, Analytic Number Theory: An Introductory Course.

In the first and third 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 $s = 1$.

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.

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 $s = 1$.

Corollary 4 (p. 123) in Hlawka, Taschner, and Schoissengeier, Geometric and Analytic Number Theory

Theorem 7.10 (p. 111) in Ayoub, An Introduction to the Analytic Theory of Numbers.

Theorem 3.4.5 (p. 134) in Jameson, The Prime Number Theorem.

Theorem 5.20 (p. 239) in Narkiewicz, The Development of Prime Number Theorey from Euclid to Hardy and Littlewood

Theorem 5.9 (p. 100) in Bateman and Diamond, Analytic Number Theory: An Introductory Course.

Corollary 4 (p. 123) in Hlawka, Taschner, and Schoissengeier, Geometric and Analytic Number Theory.

Theorem 7.10 (p. 111) in Ayoub, An Introduction to the Analytic Theory of Numbers.

Theorem 3.4.5 (p. 134) in Jameson, The Prime Number Theorem.

Theorem 5.20 (p. 239) in Narkiewicz, The Development of Prime Number Theorey from Euclid to Hardy and Littlewood.

Theorem 5.9 (p. 100) in Bateman and Diamond, Analytic Number Theory: An Introductory Course.

In the first and third 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 $s = 1$.