The series converges if and only if $p > 1$

For $p = 2$, the series converges to $\frac{\pi^2}{6}$

For $p \leq 1$, the series diverges by the Divergence Test when $p < 1$, and by the Integral Test when $p = 1$

For $p = 0.5$, the series converges because $\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0$