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单词 SumOffracmunn
释义

sum of μ(n)n


The following result holds:

n=1μ(n)n=0

where μ(n) is the Möbius function (http://planetmath.org/MoebiusFunction).

Proof:
Let n=1μ(n)n=α. Assume α0.

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

1ζ(s)=n=1μ(n)ns

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

  • 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

ζ(1)=1n=1μ(n)n=1α

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

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更新时间:2025/5/4 18:49:35