The following result holds: $$ \sum_{n=1}^{\infty} \frac{\mu(n)}{n} = 0 $$

where $\mu(n)$ is the Möbius function.

Proof:
Let $\sum_{n=1}^{\infty} \frac{\mu(n)}{n} = \alpha$ . Assume $\alpha \neq 0$ .

For $\operatorname{Re}(s) > 1$ we have the Euler product expansion $$ \frac{1}{\zeta(s)} = \sum_{n=1}^\infty \frac{\mu(n)}{n^s} $$

where $\zeta(s)$ is the Riemann zeta function.

We recall the following properties of the Riemann zeta function (which can be found in the PlanetMath entry Riemann Zeta Function).

• $\zeta(s)$ is analytic except at the point $s=1$ where it has a simple pole with residue $1$ .
• $\zeta(s)$ has no zeroes in the region $\operatorname{Re}(s) \geq 1$ .
• The function $(s-1) \zeta(s)$ is analytic and nonzero for $\operatorname{Re}(s) \geq 1$ .
• Therefore, the function $\frac{1}{\zeta(s)}$ is analytic for $\operatorname{Re}(s) \geq 1$ .

Further, as a corollary of the proof of the prime number theorem, we also know that this sum, $\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$ converges to $\frac{1}{\zeta(s)}$ for $\operatorname{Re}(s) \geq 1$ ; in particular, it converges at $s=1$ ).

But then $$ \zeta(1) = \frac{1}{\sum_{n=1}^\infty \frac{\mu(n)}{n}} = \frac{1}{\alpha} $$

So $\zeta(1)=\frac{1}{\alpha}$ , but this is a contradiction since $\zeta$ has a simple pole at $s=1$ . Therefore $\alpha = 0$ .