Answer
$${\text{The improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_1^\infty {\frac{1}{{\root 4 \of x }}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& \int_1^\infty {\frac{1}{{\root 4 \of x }}dx} = \mathop {\lim }\limits_{b \to \infty } \int_1^b {\frac{1}{{\root 4 \of x }}} dx \cr
& = \mathop {\lim }\limits_{b \to \infty } \int_1^b {{x^{ - 1/4}}} dx \cr
& {\text{Integrate using }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{x^{3/4}}}}{{3/4}}} \right]_1^b \cr
& = \frac{4}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {{b^{3/4}} - {1^{3/4}}} \right] \cr
& = \frac{4}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {{b^{3/4}} - 1} \right] \cr
& {\text{Evaluate the limit}} \cr
& = \frac{4}{3}\left( {{\infty ^{3/4}} - 1} \right) \cr
& {\text{Simplify}} \cr
& \infty \cr
& {\text{Therefore:}} \cr
& {\text{The improper integral diverges}} \cr} $$
