Answer
$${\text{diverges}}$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{{dx}}{{\sqrt x }}} \cr
& {\text{definition of improper integral}} \cr
& \int_2^\infty {\frac{{dx}}{{\sqrt x }}} = \mathop {\lim }\limits_{b \to \infty } \int_2^b {\frac{{dx}}{{\sqrt x }}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_2^b {{x^{ - 1/2}}dx} \cr
& {\text{evaluate the integral}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)} \right|_2^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left. {\left( {2{x^{1/2}}} \right)} \right|_2^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {2{b^{1/2}} - 2{{\left( 2 \right)}^{1/2}}} \right) \cr
& {\text{simplify}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left( {2{b^{1/2}} - {{\left( 2 \right)}^{3/2}}} \right) \cr
& {\text{evaluate the limit}} \cr
& = 2{\left( \infty \right)^{1/2}} - {\left( 2 \right)^{3/2}} \cr
& = \infty \cr
& {\text{diverges}} \cr} $$
