Answer
The integral converges.
Work Step by Step
Use the limit comparison test
$\lim\limits_{x \to \infty} \dfrac{1/e^x-2^x}{1/e^x}=\lim\limits_{x \to \infty} \dfrac{1}{1-(2/e)^x}=\dfrac{1}{1-0}=1$
Since we have a non-zero finite limit, the functions will either both converge or diverge. We need to test $g(x)$ for convergence.
So, $\int_1^{\infty} e^{-x} dx=\lim\limits_{a \to \infty}\int_1^{a} e^{-x} dx=\lim\limits_{a \to \infty}[-e^{-x}]_1^a=\lim\limits_{a \to \infty}[-e^{-a}-(-e^{-1})]=\dfrac{1}{e}$
Since the integral of the function $g(x)$ converges, then the original integral converges as well.