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