University Calculus: Early Transcendentals (3rd Edition)

The answer is No. Because, we have $\Sigma_{n=1}^\infty\dfrac{1}{nx}$ or, $=\dfrac{1}{x} \Sigma_{n=1}^\infty \dfrac{1}{n}$ $\Sigma_{n=1}^\infty \dfrac{1}{n}$ diverges by the n-th test for divergence. Hence, the answer is NO