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

The following result holds:

$$\sum _{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n}=0$$

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

Proof:
Let ${\sum }_{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n}=\alpha$. Assume $\alpha \ne 0$.

For $\mathrm{Re}(s)>1$ we have the Euler product expansion

$$\frac{1}{\zeta (s)}=\sum _{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n^s}$$

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 $\mathrm{Re}(s)\ge 1$.

• The function $(s-1)\zeta (s)$ is analytic and nonzero for $\mathrm{Re}(s)\ge 1$.

• Therefore, the function $\frac{1}{\zeta (s)}$ is analytic for $\mathrm{Re}(s)\ge 1$.

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

But then

$$\zeta (1)=\frac{1}{{\sum }_{n=1}^{\mathrm{\infty }}\frac{\mu (n)}{n}}=\frac{1}{\alpha }$$

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 $\frac{\mu (n)}{n}$
Canonical name	SumOffracmunn
Date of creation	2013-03-22 14:25:46
Last modified on	2013-03-22 14:25:46
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	13
Author	mathcam (2727)
Entry type	Result
Classification	msc 11A25
Related topic	MoebiusFunction