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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. 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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. 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