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