Answer
Diverges
Work Step by Step
Apply the integral test.
Here, we have $\int_1^{\infty} (\dfrac{1}{x^2}-\dfrac{1}{x}) \ dx=\lim\limits_{a \to \infty} \int_1^{a} (\dfrac{1}{x^2}-\dfrac{1}{x}) \ dx$
or, $=\lim\limits_{a \to \infty} [\dfrac{-1}{x}-\ln (x)]_1^{a}$
or, $=\lim\limits_{a \to \infty} [\dfrac{-1}{x}]-\lim\limits_{a \to \infty} [\ln (x)]$
or, $=- \infty$
Therefore, the series diverges by the integral test.
