On a sum involving the divisor function

Abstract

Let d(n) be the divisor function and denote by [t] the integral part of the real number t. In this short note, we prove that $\sum _{n\le x} \ d\Big (\Big [\frac{x}{n}\Big ]\Big ) = x\sum _{m\ge 1}\frac{\ d(m)}{m(m+1)} + O_{\varepsilon }\big (x^{11/23+\varepsilon }\big )$ for $x\rightarrow \infty$.

1 Introduction

Let [x] denote the integer part of x, i.e., $[x]=\max \{n\in {\mathbb {Z}}:n\leqslant x\}$. Let d(n) be the divisor function. From [1], we know that

$$\begin{aligned} \sum _{n\leqslant x}\left[ \frac{x}{n}\right] =x\log x+(2\gamma -1)x+O\left( x^{{517}/{1648}+o(1)}\right) , \end{aligned}$$

where $\gamma =0.577215664\cdots$ is the Euler constant. Let f be any complex-valued arithmetic function and define

$$\begin{aligned} S_{f}(x):=\sum _{n\leqslant x}f\left( \left[ \frac{x}{n}\right] \right) . \end{aligned}$$

Let $\tau _k(n)$ be the generalised divisor function, which is defined as the number of ordered representations $n=d_1\cdots d_k$ with integer numbers $d_1,\ldots ,d_k\geqslant 1$. Define $\varepsilon _1(x)=0$ and

$$\begin{aligned} \varepsilon _k(x) := \bigg (\frac{k\log _3x}{\log _2x}\bigg )^{1/2} \bigg (k-1+\frac{30}{\log _3x}\bigg ) \end{aligned}$$

(1.1)

for $k\geqslant 2$. Here $\log _r$ denotes the r-fold iterated logarithm. Bordellès, Dai, Heyman, Pan and Shparlinski (see [2, Theorem 2.6]) proved that if there exists $0<\alpha <2$ such that

$$\begin{aligned} \sum _{n\le x} |f(n)|^2 \ll x^{\alpha }, \end{aligned}$$

(1.2)

then

$$\begin{aligned} S_f(x) = x \sum _{n\ge 1} \frac{f(n)}{n(n+1)} + O\left( x^{(\alpha +1)/3} (\log x)^{(1+\alpha )(2+\varepsilon _2(x))/6}\right) , \end{aligned}$$

(1.3)

where $\varepsilon _2(x)$ is given by (1.1). A straightforward calculation shows that

$$\begin{aligned} \sum _{n\le x} d(n)^2\ll x\log ^3 x \quad (x\ge 1). \end{aligned}$$

It is easy to find that, for any $\varepsilon >0$, d(n) satisfies (1.2) with $\alpha =1+\varepsilon$. Applying (1.3), we get

$$\begin{aligned} S_{d}(x) = x\sum _{n\ge 1}\frac{d(n)}{n(n+1)} + O_{\varepsilon }\big (x^{2/3+\varepsilon }\big )\quad ( x\ge 2). \end{aligned}$$

(1.4)

In [6], Wu improved the results of Bordellès–Dai–Heyman–Pan–Shparlinski. Wu (see [6, Theorem 1.2]) proved that if there is a constant $\vartheta \in [0, 1)$ such that $|f(n)|\ll n^{\vartheta }$ for $n\ge 1$, then

$$\begin{aligned} S_f(x) = x \sum _{n\ge 1} \frac{f(n)}{n(n+1)} + O\big (x^{(\vartheta +1)/2}\big ) \quad (x\rightarrow \infty ). \end{aligned}$$

(1.5)

It is well known that $0\leqslant d(n)\ll _{\varepsilon } n^{\varepsilon }$ for $n\ge 1$ and any $\varepsilon >0$. Applying (1.5), we derive

$$\begin{aligned} S_{d}(x) = x\sum _{n\ge 1}\frac{d(n)}{n(n+1)} + O_{\varepsilon }\big (x^{1/2+\varepsilon }\big )\quad ( x\ge 2 ). \end{aligned}$$

(1.6)

This sharpens (1.4).

Several authors considered that case when f is the Euler totient function $\varphi$ (see [2, 5, 7]). In [2], Bordellès–Dai–Heyman–Pan–Shparlinski proved that

$$\begin{aligned} S_\varphi (x) \leqslant \left( \frac{2629}{4009}\cdot \frac{1}{\zeta (2)}+\frac{1380}{4009}+o(1)\right) x\log x \quad (x\geqslant 3), \\ S_\varphi (x) \geqslant \left( \frac{2629}{4009}\cdot \frac{1}{\zeta (2)}+o(1)\right) x\log x \quad (x\rightarrow \infty ). \end{aligned}$$

They also posed a question: Is it true that

$$\begin{aligned} S_\varphi (x)=\frac{x\log x }{\zeta (2)}+o(x\log x) \end{aligned}$$

(1.7)

as $x\rightarrow \infty$? In [5], Wu used the exponential sum technique to prove that

$$\begin{aligned} \frac{4}{\pi ^2}x\log x+O(x) \leqslant S_\varphi (x) \leqslant \left( \frac{1}{3}+\frac{4}{\pi ^2}\right) x\log x+O(x) \quad (x\rightarrow \infty ). \end{aligned}$$

Later, Zhai [7] proved that (1.7) is true; moreover, he got the following more precise result:

$$\begin{aligned} S_\varphi (x) = \frac{x\log x}{\zeta (2)}+O\left( x(\log x)^{2/3}(\log \log x)^{1/3}\right) \quad (x\rightarrow \infty ). \end{aligned}$$

Very recently, Ma and Wu [4] studied the asymptotic behaviour of

$$\begin{aligned} S_\Lambda (x) := \sum _{n\le x} \Lambda \Big (\Big [\frac{x}{n}\Big ]\Big ) \end{aligned}$$

and proved that

$$\begin{aligned} S_{\Lambda }(x) = x\sum _{d\ge 1}\frac{\Lambda (d)}{d(d+1)} + O_{\varepsilon }\big (x^{35/71+\varepsilon }\big ) \quad (x\rightarrow \infty ). \end{aligned}$$

(1.8)

This result is better than ones can be derived applying theorems from [2, 6]. In [4], using the Vaughan identity (see [4, Lemma 2.1]) and the exponential pair method, Ma and Wu gave a non-trivial bound of

$$\begin{aligned} \sum _{D<d\leqslant {}2D}\Lambda (d)\psi \left( \frac{x}{d+\delta }\right) , \end{aligned}$$

which plays a key role in the proof of (1.8).

Inspired by [4], we study the summation $S_d$ and obtain a better estimate than (1.6). Let $\psi (t):=t-[t]-\frac{1}{2}$ and $\delta \in \{0,1\}.$ For $x\geqslant {}2$ and  $1\leqslant {} D \leqslant {} x ,$ define

$$\begin{aligned} {\mathfrak {S}}_{\delta }(x,D):=\sum _{D<k\leqslant {}2D}d(k)\psi \left( \frac{x}{k+\delta }\right) . \end{aligned}$$

Noticing the expression $d(k)=\sum _{mn=k}{\mathbf {1}}(m){\mathbf {1}}(n)$ and the symmetry of the factors m and n, we turn ${\mathfrak {S}}_{\delta }(x,D)$ into a special Type I sum (see [3, Chapter 4, Page 50]) and eventually get the following theorem.

Theorem 1.1

For any $\varepsilon >0$, we have

$$\begin{aligned} S_{d}(x) = x\sum _{m\ge 1}\frac{d(m)}{m(m+1)} + O_{\varepsilon }\big (x^{11/23+\varepsilon }\big ) \end{aligned}$$

as $x\rightarrow \infty$. We note that $\sum _{m\geqslant 1}\frac{d(m)}{m(m+1)}\approx 1.88$.

2 A key estimate

In this section, we will prove the following bound for ${\mathfrak {S}}_{\delta }(x,D)$, which plays a key role in the proof of Theorem 1.1. We refer to [3, Chapter 3, page 31] for the definition and basic property of exponent pairs.

Proposition 2.1

Under the previous notation, we have

$$\begin{aligned} \begin{aligned} {\mathfrak {S}}_{\delta }(x,D)\ll \left( x^{2\kappa }D^{\lambda -\kappa +1}(\log D)^{2\kappa }\right) ^{{1}/{(2\kappa +2)}}+x^{\kappa } D^{(\lambda -3\kappa +1)/2}+x^{-1}D^2\log D, \end{aligned} \end{aligned}$$

(2.1)

where $(\kappa ,\lambda )$ is an exponent pair. In particular, we have

$$\begin{aligned} \begin{aligned} {\mathfrak {S}}_{\delta }(x,D)\ll x^{1/7}D^{9/14}(\log D)^{1/7}+x^{1/6} D^{7/12}+x^{-1}D^2\log D. \end{aligned} \end{aligned}$$

(2.2)

2.1 An important lemma

In order to prove Proposition 2.1, we need a result of Vaaler (see [3, Theorem A.6]).

Lemma 2.2

For $x\geqslant {}1$ and  $H\geqslant {}1$, we have

$$\begin{aligned} \psi (x)=-\sum _{1\leqslant {} |h|\leqslant {} H} \Phi \left( \frac{h}{H+1}\right) \frac{e(hx)}{2\pi ih}+R_H(x), \end{aligned}$$

where $e(t):=e^{2\pi it},\Phi (t):=\pi t(1-|t|)\cos (\pi t)+|t|$, and the error term  $R_H(x)$ satisfies

$$\begin{aligned} |R_H(x)|\leqslant {}\frac{1}{2H+2}\sum _{|h|\leqslant {} H}\left( 1-\frac{|h|}{H+1}\right) e(hx). \end{aligned}$$

2.2 Proof of Proposition 2.1

2.2.1 The initial transformation

For an integer $n\geqslant 1$, let d(n) denote the divisor function. By using the relation

$$\begin{aligned} d(k)=\sum _{mn=k}{\mathbf {1}}(m){\mathbf {1}}(n) \end{aligned}$$

and the symmetry of factors, we have

$$\begin{aligned} \begin{aligned} {\mathfrak {S}}_{\delta }(x,D)&=\sum _{D< k\leqslant {} 2D}d(k)\psi \left( \frac{x}{k+\delta }\right) \\&=\sum _{1\leqslant {} m\leqslant {} 2D}\sum _{\frac{D}{m}<n\leqslant {} \frac{2D}{m}}\psi \left( \frac{x}{mn+\delta }\right) \\&=2\sum _{m\leqslant \sqrt{2D}} \sum _{\frac{D}{m}<n\leqslant {} \frac{2D}{m}}\psi \left( \frac{x}{mn+\delta }\right) . \end{aligned} \end{aligned}$$

By Lemma 2.2 we derive that

$$\begin{aligned} \begin{aligned} {\mathfrak {S}}_{\delta }(x,D)=&-\frac{1}{\pi i}\sum _{m\leqslant \sqrt{2D}} \sum _{\frac{D}{m}<n\leqslant {} \frac{2D}{m}}\sum _{1\leqslant {} |h|\leqslant {} H}\frac{1}{h} \Phi \left( \frac{h}{H+1}\right) e\left( \frac{hx}{mn+\delta }\right) \\&\qquad +2\sum _{m\le \sqrt{2D}}\sum _{D<mn\le 2D} R_H\Big (\frac{x}{mn+\delta }\Big ). \end{aligned} \end{aligned}$$

Now we write

$$\begin{aligned} {\mathfrak {S}}_{\delta }(x,D) = - \frac{1}{\pi \text {i}} \big ({\mathfrak {S}}^{\sharp }_{\delta } + \overline{{\mathfrak {S}}^{\sharp }_{\delta }}\big ) + {\mathfrak {S}}^{\dag }_{\delta }, \end{aligned}$$

(2.3)

where $H\ge 1$ and

$$\begin{aligned} {\mathfrak {S}}^{\sharp }_{\delta }&:= \sum _{m\le \sqrt{2D}}\sum _{D<mn\le 2D} \sum _{h\le H}\frac{1}{h} \Phi \Big (\frac{h}{H+1} \Big ) \text {e}\Big (\frac{hx}{mn+\delta } \Big ), \\ {\mathfrak {S}}^{\dagger }_{\delta }&:= 2\sum _{m\le \sqrt{2D}}\sum _{D<mn\le 2D} R_H\Big (\frac{x}{mn+\delta }\Big ). \end{aligned}$$

2.2.2 A bound of ${\mathfrak {S}}^{\sharp }_{\delta }$

Noticing the fact that $0<\Phi (t)<1$ $(0<|t|<1)$ and applying the exponent pair $(\kappa ,\lambda )$ to the sum over n, we derive

$$\begin{aligned} \begin{aligned} {\mathfrak {S}}^{\sharp }_{\delta }&\ll \sum _{m\le \sqrt{2D}} \sum _{h\le H} \frac{1}{h} \bigg \{\Big (\frac{hx}{(D^2/m)}\Big )^{\kappa } \Big (\frac{D}{m}\Big )^{\lambda } + \frac{(D^2/m)}{hx}\bigg \} \\&\ll \sum _{m\le \sqrt{2D}} \big (x^{\kappa } D^{-2\kappa +\lambda } m^{\kappa -\lambda } H^{\kappa } + x^{-1} D^2m^{-1}\big ) \\&\ll x^{\kappa } D^{(-3\kappa +\lambda +1)/2} H^{\kappa } + x^{-1}D^2\log D \end{aligned} \end{aligned}$$

(2.4)

for all $H\ge 1$.

2.2.3 A bound of ${\mathfrak {S}}^{\dag }_{\delta }$

It is clear that

$$\begin{aligned} \big |{\mathfrak {S}}^{\dagger }_{\delta }\big |&\ll \sum _{m\le \sqrt{2D}} \sum _{D<mn\le 2D} \Big |R_H\Big (\frac{x}{mn+\delta }\Big )\Big | \ll DH^{-1}\log D + \big |\widetilde{{\mathfrak {S}}}^{\dagger }_{\delta }\big | , \end{aligned}$$

where

$$\begin{aligned} \widetilde{{\mathfrak {S}}}^{\dagger }_{\delta } := \frac{1}{H} \sum _{m\le \sqrt{2D}} \sum _{D<mn\le 2D} \sum _{1\le |h|\le H} \Big (1-\frac{|h|}{H+1}\Big ) \mathrm{e}\Big (\frac{hx}{mn+\delta }\Big ). \end{aligned}$$

Now, similarly to the case of ${\mathfrak {S}}^{\sharp }_{\delta }$, for all $H\ge 1$, we obtain that

$$\begin{aligned} {\mathfrak {S}}^{\dagger }_{\delta } \ll D H^{-1}\log D + x^{\kappa } D^{(-3\kappa +\lambda +1)/2} H^{\kappa } + x^{-1} D^2\log D. \end{aligned}$$

(2.5)

2.2.4 Concluding the proof

Combining (2.5) and (2.4) with (2.3) gives

$$\begin{aligned} {\mathfrak {S}}_{\delta }(x,D) \ll D H^{-1}\log D + x^{\kappa } D^{(-3\kappa +\lambda +1)/2} H^{\kappa } + x^{-1} D^2\log D \end{aligned}$$

for $H\ge 1$. Optimising H over $[1, \infty )$, it follows that

$$\begin{aligned} {\mathfrak {S}}_{\delta }(x,D) \ll \left( x^{2\kappa } D^{-\kappa +\lambda +1}(\log D)^{2\kappa }\right) ^{1/(2\kappa +2)} + x^{\kappa } D^{(-3\kappa +\lambda +1)/2} + x^{-1} D^2\log D. \end{aligned}$$

In (2.1), taking $(\kappa ,\lambda )=AB(0,1)=(\frac{1}{6},\frac{2}{3})$ , we get the desired estimate (2.2).

3 Proof of Theorem 1.1

Let $N\in [1, x^{1/2})$ be a parameter that will be choosen later. The sum $S_d(x)$ can be split in two parts:

$$\begin{aligned} S_d(x) :=S_1(x)+S_2(x) \end{aligned}$$

(3.1)

where

$$\begin{aligned} S_1(x):=\sum _{n\le N}d\left( \left[ \frac{x}{n}\right] \right) , \qquad S_2(x):=\sum _{N<n\le x}d\left( \left[ \frac{x}{n}\right] \right) . \end{aligned}$$

A. A bound of $S_1(x)$

Using that $d(n)\ll _{\varepsilon } n^{\varepsilon }$ for all $n\ge 1$ and any $\varepsilon >0$, we have

$$\begin{aligned} \begin{aligned} S_1(x) \ll x^{\varepsilon }\sum _{n\leqslant {} N}\frac{1}{n^{\varepsilon }} \ll N\left( \frac{x}{N}\right) ^{\varepsilon } \ll Nx^{\varepsilon }. \end{aligned} \end{aligned}$$

(3.2)

B. A bound of $S_2(x)$

In order to get a bound of $S_2(x)$, put $m=[{x}/{n}]$. Then

$$\begin{aligned} {x}/{n}-1<m \leqslant {} {x}/{n} \qquad \text {and} \qquad {x}/{(m+1)}<n\leqslant {} {x}/{m}. \end{aligned}$$

Thus we have

$$\begin{aligned} \begin{aligned} S_2(x)&=\sum _{N<n\leqslant {} x}d\left( \left[ \frac{x}{n}\right] \right) \\&=\sum _{m\leqslant {} {x}/{N}}d(m)\sum _{{x}/{(m+1)}<n \leqslant {} {x}/{m}}1\\&=\sum _{m\leqslant {} {x}/{N}}d(m)\left( \frac{x}{m}-\psi \left( \frac{x}{m}\right) -\frac{x}{m+1}+\psi \left( \frac{x}{m+1}\right) \right) \\&=x\sum _{m\leqslant {} {x}/{N}}\frac{d(m)}{m(m+1)}+ \sum _{m\leqslant {} {x}/{N}}d(m)\left( \psi \left( \frac{x}{m+1}\right) -\psi \left( \frac{x}{m}\right) \right) . \end{aligned} \end{aligned}$$

It is well known that

$$\begin{aligned} d(n)=O(n^{\varepsilon }),\qquad \qquad \sum _{n\leqslant x} d(n)=x\log x+(2\gamma -1)x+O(x^{\frac{1}{2}}). \end{aligned}$$

With the help of these, we can derive

$$\begin{aligned} x\sum _{m> x/N}\frac{d(m)}{m(m+1)}\ll N^{1+\varepsilon }, \qquad \sum _{m\le N}d(m)\Big (\psi \Big (\frac{x}{m+1}\Big ) - \psi \Big (\frac{x}{m}\Big )\Big ) \ll N^{1+\varepsilon }. \end{aligned}$$

So we have

$$\begin{aligned} S_2(x)=x\sum _{m\geqslant {} 1}\frac{d(m)}{m(m+1)}+R_{1}(x) - R_{0}(x) + O(N^{1+\varepsilon }), \end{aligned}$$

(3.3)

where

$$\begin{aligned} R_{\delta }(x)=\sum _{N< m\le x/N}d(m) \psi \Big (\frac{x}{m+\delta }\Big ). \end{aligned}$$

Motivated by [5], let  $D_k:={x}/({2^kN}),$ and let K be the integer such that  $D_{K+1}<N\leqslant {} D_K$. By a simple dyadic split and by Proposition 2.1 with  $(\kappa ,\lambda )=(\frac{1}{6},\frac{2}{3})$, it follows that

$$\begin{aligned} \begin{aligned} |R_{\delta }(x)|&\leqslant {} \sum _{1\leqslant {} k\leqslant {} K+1}|{\mathfrak {S}}_{\delta }(x,D_k)|\\&\ll \sum _{1\leqslant {} k\leqslant {} K+1}(x^{1/7}D_{k}^{9/14}(\log D_{k})^{1/7}+x^{1/6}D_{k}^{7/12}+x^{-1}D_k^2\log D_k)\\&\ll x^{1/7}{(x/N)}^{9/14}\left( \log ({x/N})\right) ^{1/7}+x^{1/6}{(x/N)}^{7/12}+x^{-1}({x}/{N})^2\log ( {x}/{N})\\&\ll x^{11/14}N^{-9/14}(\log ({x/N}))^{1/7}+x^{9/12}N^{-7/12}+xN^{-2}\log ({x}/{N}). \end{aligned} \end{aligned}$$

Inserting this estimate into (3.3), we obtain

$$\begin{aligned} S_2(x) = x\sum _{m\ge 1} \frac{d(m)}{m(m+1)} +R(x), \end{aligned}$$

(3.4)

where

$$\begin{aligned} R(x)=O\left( N^{1+\varepsilon }+x^{11/14}N^{-9/14}(\log ({x/N}))^{1/7}+x^{9/12}N^{-7/12}+xN^{-2}\log ({x}/{N})\right) . \end{aligned}$$

C. Concluding the proof

Combining (3.2) and (3.4) with (3.1), we obtain

$$\begin{aligned} S_d(x)=x\sum _{m\ge 1} \frac{d(m)}{m(m+1)}+{\widetilde{R}}(x), \end{aligned}$$

where

$$\begin{aligned} {\widetilde{R}}(x)=O\left( Nx^{\varepsilon }+N^{1+\varepsilon }+x^{11/14}N^{-9/14}(\log ({x/N}))^{1/7}+x^{9/12}N^{-7/12}+xN^{-2}\log ({x}/{N})\right) . \end{aligned}$$

Choosing $N=x^{11/23}$, we have

$$\begin{aligned} S_{d}(x) = x\sum _{m\ge 1}\frac{d(m)}{m(m+1)} + O_{\varepsilon }\big (x^{11/23+\varepsilon }\big ). \end{aligned}$$

This completes the proof.