# Is this a valid z-transform of $\frac{1}{n^2}u[n]$?

I was looking over some DSP and came across the following signal: $x[n] = 1/n$. So I wondered whether it had a z-transform but I soon realized that it fails to meet two conditions described here:

...(1) the signal starts at a finite time and (2) it is asymptotically exponentially bounded...

So I modified it to meet those requirements: $x[n] = \frac{1}{n^2} u[n]$

(I assume $n>0$ is a condition or perhaps $u[n]$ should be changed to $u[n-1]$.)

Considering that signal, I believe it satisfies both requirements however I don't see anything like it in a table of common z-transforms, but it can be rewritten as:

$x[n] = \frac{1}{1^2}*\delta[n-1] + \frac{1}{2^2}*\delta[n-2] + \frac{1}{3^2}*\delta[n-3]\ +\ ...$

Which I believe transforms to $X(z) = \frac{1}{z^1} + \frac{1}{(2z)^2} + \frac{1}{3^2z^3} +\ ...$

Does this transformation make sense? Is it the best approach? And if it's correct, is there a more compact way to express it?

• The Z-transform of $\frac{1}{n^2} u[n-1]$ is explained in this answer. Jul 20, 2016 at 14:48

What you've got is definitely the z-transform of that signal (by definition). For a more compact expression, notice that the polylogarithm is exactly the infinite sum you've shown (just replace $z \rightarrow 1/z$ in the Mathworld expression). Unfortunately I don't think there's any nicer way to write it; sometimes there just isn't.