Answer
Convergent
Work Step by Step
$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\dfrac{(2+cosn)a_{n}}{\sqrt n}}{a_{n}}|$
$=\lim\limits_{n \to \infty}\frac{2+cosn}{\sqrt n}$
$=0$
The $(2+cosn)$ oscillates between $1$ and $3$ while $\sqrt n$ goes to $\infty$ . The limit is $\lt 1$, so it absolutely converges (and thus is convergent).