Answer
Diverges.
Work Step by Step
Use the limit comparison test.
$\lim\limits_{x \to \infty} \dfrac{\dfrac{1}{\sqrt {x}}-1}{\dfrac{1}{\sqrt {x}}}=\lim\limits_{x \to \infty} \dfrac{\sqrt x}{\sqrt {x}-1}$
Use L'Hospital's Rule:
$ \lim\limits_{x \to \infty} \dfrac{\dfrac{1}{2 \sqrt {x}}}{1/2\sqrt x}=1$
and $\lim\limits_{a \to \infty}\int_{4}^{a} x^{-(1/2)} dx=\lim\limits_{a \to \infty}[2 \sqrt x]_{4}^a\\=\lim\limits_{a \to \infty}[2\sqrt a-4] \\=\infty$
So, the integral diverges.