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 $s = 1$.