Answer
$$Convergent,1-\frac{1}{e}$$
Work Step by Step
$let\ t =-\frac{1}{x},\ dt=\frac{1}{x^{2}}dx$
$\int_{1}^{\infty}\Rightarrow \int_{-1}^{0}$
$$\int_{1}^{\infty}\frac{e^{-1/x}}{x^{2}}dx=\int_{-1}^{0}e^{t}dt$$
$$=\left [ e^{t} \right ]_{-1}^{0}=1-\frac{1}{e}$$
Convergent.
