Note on a sum involving the divisor function
Liuying Wu
TL;DR
This paper studies the sum $S_{d,c}(x)=\sum_{n\le x^{1/c}} d\left(\left[\frac{x}{n^{c}}\right]\right)$, establishing an improved asymptotic expansion compared to Feng. The authors combine a Vaaler-type approximation for the sawtooth function with a Jutila-type exponential-sum bound to control oscillatory terms arising from floor functions, and carefully optimize auxiliary parameters to balance main-term extraction against error accumulation. The main result is $S_{d,c}(x)=x^{1/c}\sum_{k\ge1} d(k)\left(\frac{1}{k^{1/c}}-\frac{1}{(k+1)^{1/c}}\right)+O_{\varepsilon,c}\left(x^{\theta_c+\varepsilon}\right)$, where $\theta_c=2c+22c^2+5c+2$ for $0<c<\tfrac{2}{3}$ and $\theta_c=55c+6$ for $c\ge\tfrac{2}{3}$. This improves Feng's exponent for $c>\tfrac{2}{9}$ and generalizes prior work (e.g., $c=1$ with $\theta_1=5/11$). The approach highlights the power of combining precise Fourier-analytic approximations with selective exponential-sum bounds in divisor-function problems involving floor constraints.
Abstract
Let $d(n)$ be the divisor function and denote by $[t]$ the integral part of the real number $t$. In this paper, we prove that $$\sum_{n\leq x^{1/c}}d\left(\left[\frac{x}{n^c}\right]\right)=d_cx^{1/c}+\mathcal{O}_{\varepsilon,c} \left(x^{\max\{(2c+2)/(2c^2+5c+2),5/(5c+6)\}+\varepsilon}\right),$$ where $d_c=\sum_{k\geq1}d(k)\left(\frac{1}{k^{1/c}}-\frac{1}{(k+1)^{1/c}}\right)$ is a constant. This result constitutes an improvement upon that of Feng.