Answer
Converges
Work Step by Step
Since, $\lim\limits_{t \to \infty} \dfrac{2/t^{3/2}-1}{2/t^{3/2}}=\lim\limits_{t \to \infty} \dfrac{t^{3/2}}{t^{3/2}-1}=1$
As the limit is finite, then $$\int_4^{\infty} \dfrac{2 dt}{t^{3/2}}=\lim\limits_{a \to \infty}\int_4^{a} \dfrac{2 dt}{t^{3/2}} \\= \\=\lim\limits_{a \to \infty}[-4/\sqrt x]_4^{a} \\=\lim\limits_{a \to \infty}[-4/\sqrt a-(-\dfrac{4}{\sqrt 4})]= \\=0+2\\= 2$$
Thus, the integral converges by the limit comparison test.