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The Dirichlet series of Euler's totient is zeta of s minus 1 divided by zeta of s on Re s greater than 2

Statement

For $ℜ s > 2$ ,

$\sum_{n \geq 1} φ (n) n^{- s} = \frac{ζ (s - 1)}{ζ (s)},$

where $ζ (s) = \sum_{n \geq 1} n^{- s}$ .

Facts & Assumptions

Given: A complex number $s$ with $ℜ s > 2$ .

[L1]

Euler's totient satisfies $\sum_{d ∣ n} φ (d) = n$ , where $n = {id}_{1} (n)$ (For every positive integer $n$ , $\sum_{d ∣ n, d > 0} φ (d) = n$ , The power functions ${id}_{k}$ and the divisor-power-sum functions $σ_{k}$ , The unit group $(Z / n)^{\times}$ and Euler's totient $φ (n) = ∣ (Z / n)^{\times} ∣$ for $n \geq 1$ ).

[L2]

Classical Möbius inversion and Dirichlet-series multiplication convert that identity into convolution identities (Classical Möbius inversion over positive divisors, Multiplying absolutely convergent Dirichlet series gives Dirichlet convolution).

[L3]

The Möbius Dirichlet series is $1 / ζ (s)$ (The Dirichlet series of the Möbius function is the reciprocal of the zeta Dirichlet series on Re s greater than 1).

[L4]

For every rational $q > 1$ , the series $\sum_{n \geq 1} n^{- q}$ converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

Proof

technique · direct

1.1L1L2givenalgebra

By [L1] and Möbius inversion from [L2], $φ = μ * {id}_{1} .$

2.1L4step 1.1givenchoosealgebra

Write $σ : = ℜ s > 2$ and choose a rational $q$ with $1 < q < σ - 1$ . Since $μ (n) \in {- 1, 0, 1}$ , one has $∣ μ (n) n^{- s} ∣ \leq n^{- σ} \leq n^{- q}, ∣ n^{1 - s} ∣ = n^{1 - σ} \leq n^{- q} .$ By [L4], both Dirichlet series in the next step converge absolutely.

3.1step 1.1step 2.1algebra

Therefore Multiplying absolutely convergent Dirichlet series gives Dirichlet convolution gives $\sum_{n \geq 1} φ (n) n^{- s} = (\sum_{n \geq 1} μ (n) n^{- s}) (\sum_{n \geq 1} n^{1 - s}) .$

4.1L3step 3.1∎

The first factor is $1 / ζ (s)$ by [L3], and the second is $ζ (s - 1)$ by definition. Hence the product is $ζ (s - 1) / ζ (s)$ .

Depends on

Used by

A local prime factor in the Dirichlet series of Euler's totient Example

Dependency tree · two levels

44 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Kiran S. Kedlaya, Notes on Analytic Number Theory, Chapter 2 examples (standard reference, not scraped)