Answer
$\frac{1}{{3\ln 3}}$
Work Step by Step
$$\eqalign{
& \int_1^\infty {{{\left( {\frac{1}{3}} \right)}^x}} dx \cr
& {\text{Using the definition of improper integrals }} \cr
& \underbrace {\int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{t \to \infty } } \int_a^t {f\left( x \right)} dx}_ \Downarrow \cr
& \int_1^\infty {{{\left( {\frac{1}{3}} \right)}^x}} dx = \mathop {\lim }\limits_{t \to \infty } \int_1^t {{{\left( {\frac{1}{3}} \right)}^x}} dx \cr
& {\text{Integrating}}{\text{, use }}\int {{a^x}} dx = \left( {\frac{1}{{\ln a}}} \right){a^x} + C \cr
& = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{\ln \left( {1/3} \right)}}{{\left( {\frac{1}{3}} \right)}^x}} \right]_1^t \cr
& = - \frac{1}{{\ln 3}}\mathop {\lim }\limits_{t \to \infty } \left[ {{{\left( {\frac{1}{3}} \right)}^x}} \right]_1^t \cr
& = - \frac{1}{{\ln 3}}\mathop {\lim }\limits_{t \to \infty } \left[ {{{\left( {\frac{1}{3}} \right)}^t} - {{\left( {\frac{1}{3}} \right)}^1}} \right] \cr
& = - \frac{1}{{\ln 3}}\left[ {{{\left( {\frac{1}{3}} \right)}^{\mathop {\lim }\limits_{t \to \infty } t}} - \mathop {\lim }\limits_{t \to \infty } {{\left( {\frac{1}{3}} \right)}^1}} \right] \cr
& {\text{Evaluate the limit when }}t \to \infty \cr
& = - \frac{1}{{\ln 3}}\left[ {0 - \frac{1}{3}} \right] \cr
& = \frac{1}{{3\ln 3}} \cr
& {\text{Therefore}}{\text{, the integral converges}} \cr} $$
