$\mathcal{Z}$-transform of $\frac{1}{n^2}$
The problem is not sufficiently specified, because the range of admissible values of $n$ is missing. Here I make the assumption that we consider $n&gt;0$. With this assumption we have

$$X(z)=\sum_{n=1}^{\infty}x[n]z^{-n}=\sum_{n=1}^{\infty}\frac{z^{-n}}{n^2}\tag{1}$$

And that's the point where we might get stuck, if we didn't have a list of mathematical series, or if we didn't know about the polylogarithm, which is defined by

$$\text{Li}_{s}(z)=\sum_{n=1}^{\infty}\frac{z^n}{n^s},\qquad |z|&lt;1\tag{2}$$

where $s$ is an arbitrary complex number.

In your case, $s=2$ and the corresponding function is called the dilogarithm or Spence's function.

Comparing $(1)$ and $(2)$ we get for the $\mathcal{Z}$-transform of $1/n^2$ for $n&gt;0$

$$X(z)=\text{Li}_2\left(\frac{1}{z}\right),\qquad |z|&gt;1\tag{3}$$

Another way to arrive at the solution is to use the differentiation property of the $\mathcal{Z}$-transform:

$$nx[n]\Longleftrightarrow -z\frac{dX(z)}{dz}\tag{4}$$

Applying $(4)$ twice will give you the result. You need the correspondence

$$u[n-1]\Longleftrightarrow \frac{1}{z-1},\qquad |z|&gt;1\tag{5}$$

In a first step you'll arrive at the transform of $1/n$, $n&gt;0$, and in a second step you'll arrive at the transform of $1/n^2$, $n&gt;0$.

In this case you will be using the integral representation of the dilogarithm:

$$\text{Li}_2(z)=-\int_0^z\frac{\ln(1-u)}{u}du\tag{6}$$