sum of μ(n)n

The following result holds:


where μ(n) is the Möbius function (

Let n=1μ(n)n=α. Assume α0.

For Re(s)>1 we have the Euler productMathworldPlanetmath expansion


where ζ(s) is the Riemann zeta functionMathworldPlanetmath.

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

  • ζ(s) is analytic except at the point s=1 where it has a simple pole with residue 1.

  • ζ(s) has no zeroes in the region Re(s)1.

  • The function (s-1)ζ(s) is analytic and nonzero for Re(s)1.

  • Therefore, the function 1ζ(s) is analytic for Re(s)1.

Further, as a corollary of the proof of the prime number theoremMathworldPlanetmath, we also know that this sum, n=1μ(n)ns converges to 1ζ(s) for Re(s)1; in particular, it converges at s=1).

But then


So ζ(1)=1α, but this is a contradictionMathworldPlanetmathPlanetmath since ζ has a simple pole at s=1. Therefore α=0.

Title sum of μ(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