Answer
Converges
Work Step by Step
Apply the limit comparison test.
$\lim\limits_{x \to \infty} \dfrac{\dfrac{1}{e^x}-2^x}{1/e^x}=\lim\limits_{x \to \infty} \dfrac{1}{1-(\dfrac{2}{e})^x}=\dfrac{1}{1-0}=1$
Now, $\int_1^{\infty} e^{-x} dx=\lim\limits_{a \to \infty}\int_1^{a} e^{-x} dx \\=\lim\limits_{a \to \infty}[-e^{-x}]_1^a \\=\lim\limits_{a \to \infty}[-e^{-a}-(-e^{-1})] \\=\dfrac{1}{e}$
Thus, the integral converges.