Answer
Divergent
Work Step by Step
$\int_1^\infty f(x)dx=\int_1^\infty \frac{1}{\sqrt {x+4}}dx$
$=\lim\limits_{t \to \infty}\int_1^t \frac{2}{2\sqrt {x+4}}dx$
$=\lim\limits_{t \to \infty}[ 2\sqrt {x+4}]_1^t$
$=\infty -2\sqrt 5$
$=\infty $
Hence, the given series is divergent.