Answer
Converges
Work Step by Step
Let $u_n=\dfrac{\cos^2 n}{(n^{3/2})}$
When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing.
But $-1 \leq \cos n \leq 1; 0 \leq \cos^2 n \leq 1$
$\implies u_n \leq \dfrac{1}{n^{3/2}}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{n^{3/2}}$ a convergent $p$-series with $p=\dfrac{3}{2} \gt 1$
Hence, the series converges by the comparison test.