Answer
Converges
Work Step by Step
Use the limit comparison test.
Since, $ \lim\limits_{x \to \infty} \dfrac{\sqrt{x+1}/x^2}{ 1/x^{3/2}}=\lim\limits_{x \to \infty} \dfrac{x^{3/2}\sqrt{x+1}}{ x^2}$
Thus, $\lim\limits_{x \to \infty} \dfrac{(x)^{3/2}\sqrt{x+1}}{ x^2}=\lim\limits_{x \to \infty} \sqrt {1+(\dfrac{1}{x})}=1$
Now,
$$\int_1^{\infty} \dfrac{dx}{x^{3/2}}=\lim\limits_{a \to \infty}\int_1^{a} \dfrac{dx}{x^{(3/2)}}\\ \lim\limits_{a \to \infty}\int_1^{a} \dfrac{dx}{x^{3/2}} \\=\lim\limits_{a \to \infty}[\dfrac{-2}{\sqrt x}]_1^{a} \\=-2$$
Thus, the integral converges by the limit comparison test.