Answer
No
Work Step by Step
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
