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The summatory divisor-counting function is x log x plus (2 gamma - 1)x plus O(sqrt x)

Statement

For every real $x \geq 1$ ,

$\sum_{n \leq x} τ (n) = x \log x + (2 γ - 1) x + O (\sqrt{x}) .$

Facts & Assumptions

Given: A real $x \geq 1$ and $N : = ⌊ \sqrt{x} ⌋$ .

Proof

technique · direct

1.1givenalgebra

By The divisor functions arise by Dirichlet convolution, $τ = 1 * 1$ . Applying Dirichlet's hyperbola method for summatory convolutions with $f = g = 1$ and $U = V = \sqrt{x}$ gives $\sum_{n \leq x} τ (n) = 2 \sum_{a \leq \sqrt{x}} ⌊ \frac{x}{a} ⌋ - ⌊ \sqrt{x} ⌋^{2} = 2 \sum_{a = 1}^{N} ⌊ \frac{x}{a} ⌋ - N^{2} .$

2.1step 1.1givenalgebra

Since $⌊ x / a ⌋ = x / a + O (1)$ uniformly in $a$ , summing over $1 \leq a \leq N$ yields $\sum_{a = 1}^{N} ⌊ \frac{x}{a} ⌋ = x \sum_{a = 1}^{N} \frac{1}{a} + O (N) .$

3.1step 2.1givenalgebra

By The harmonic sum is log x plus gamma plus O(1/x), $\sum_{a = 1}^{N} 1 / a = \log N + γ + O (1 / N)$ . Also $N^{2} \leq x < (N + 1)^{2}$ , so $x = N^{2} + O (N)$ and $\log N = \frac{1}{2} \log x + O (1 / N)$ . Substituting these into step 2.1 gives $2 \sum_{a = 1}^{N} ⌊ \frac{x}{a} ⌋ = x \log x + 2 γ x + O (\sqrt{x}) .$

4.1step 1.1step 3.1givenalgebra∎

Since $N^{2} = x + O (\sqrt{x})$ , combining step 3.1 with step 1.1 yields $\sum_{n \leq x} τ (n) = x \log x + (2 γ - 1) x + O (\sqrt{x}) .$

Depends on

Used by

The average order of tau is log n Corollary
The divisor summatory estimate through several small values Example

Dependency tree · two levels

16 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

Tom Sanders, Topics in Analytic Number Theory, Chapter 1 (standard reference, not scraped)
Karl-Dieter Crisman, Number Theory: In Context and Interactive (standard reference, not scraped)