Type of singularity of the Laurent series $\frac{1}{z^{2n}}$

Question

What is the type of singularity of the Laurent series $\sum_2^{\infty} 1/z^{2n}$ at point $z=0$? is it an essential singularity because the series has infinitely many terms or is it a pole because $\sum_2^{\infty} 1/z^{2n}=1/(z^2(z^2-1))$

Answer 1

$\sum \frac1 {z^{2n}}$ is divergent for $|z|\leq 1$. So it is not even defined in any neighborhood of $0$ and the type of singularity at $0$ is not defined.

However $\frac 1 {z^{2}(z^{2}-1)}$ has a pole of order $2$ at $0$.