Answer
$$\int_1^\infty {\frac{1}{{{x^{0.999}}}}} dx{\text{ is divergent}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{{x^{0.999}}}}} dx \cr
& {\text{by the definition of an improper integral}}{\text{,}} \cr
& \int_1^\infty {\frac{1}{{{x^{0.999}}}}} dx = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{{x^{0.999}}}}} dx \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_1^b {{x^{ - 0.999}}} dx \cr
& {\text{integrating by using the power rule}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{x^{ - 0.999 + 1}}}}{{ - 0.999 + 1}}} \right]_1^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{x^{1/1000}}}}{{1/1000}}} \right]_1^b \cr
& = 1000\mathop {\lim }\limits_{b \to \infty } \left[ {{x^{1/1000}}} \right]_1^b \cr
& {\text{ using ftc}} \cr
& = 1000\mathop {\lim }\limits_{b \to \infty } \left[ {{b^{1/1000}} - {1^{1/1000}}} \right] \cr
& = 1000\mathop {\lim }\limits_{b \to \infty } \left[ {{b^{1/1000}} - 1} \right] \cr
& {\text{evaluating the limit when }}b \to \infty \cr
& = 1000\left[ {{{\left( \infty \right)}^{1/1000}} - 1} \right] \cr
& = 1000\left( \infty \right) \cr
& = \infty \cr
& {\text{then}} \cr
& \int_1^\infty {\frac{1}{{{x^{0.999}}}}} dx{\text{ is divergent}} \cr} $$
