Answer
Converges
Work Step by Step
Apply the limit comparison test.
Thus, we have $ \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}$
This suggests that $\lim\limits_{x \to \infty} \dfrac{x^{3/2}\sqrt{x+1}}{ x^2}=\lim\limits_{x \to \infty} \sqrt {1+(1/x)}=1$
Since we got a finite positive value, we know that both series converge or both diverge.
$\int_1^{\infty} \dfrac{dx}{x^{3/2}}=\lim\limits_{p \to \infty}\int_1^{p} \dfrac{dx}{x^{3/2}}$
and $\lim\limits_{p \to \infty}\int_1^{p} \dfrac{dx}{x^{3/2}}=\lim\limits_{p \to \infty}[\dfrac{-2}{x^{1/2}}]_1^{p} =-2$
Hence, the given integral converges by the limit comparison test.
