# sum of $\frac{\mu(n)}{n}$

The following result holds:

 $\sum_{n=1}^{\infty}\frac{\mu(n)}{n}=0$

where $\mu(n)$ is the Möbius function (http://planetmath.org/MoebiusFunction).

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 (http://planetmath.org/RiemannZetaFunction)).

• $\zeta(s)$ 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 $\alpha=0$.

Title sum of $\frac{\mu(n)}{n}$ SumOffracmunn 2013-03-22 14:25:46 2013-03-22 14:25:46 mathcam (2727) mathcam (2727) 13 mathcam (2727) Result msc 11A25 MoebiusFunction