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Evaluate the Z transform of $x[n] = n^3$ where the signal is two sided.
I have tried using the basic definition of the Z transform ie.,
$$X(z) \triangleq \sum_\limits{n=-\infty}^{+\infty} x[n] \, z^{-n} $$
But what I get obviously is an infinite unbounded sum of sequence. Does that mean Z transform does not exist? IS there any alternate methods to obtain the transform? Please help.
[My path was erroneous, I made a confusion, but kept a modified answer for one-sided $\mathcal{Z}$-transforms]
With $\mathcal{Z}$-transforms, formulae with powers of $n$ often aim at making you handle basic operations on $\mathcal{Z}$-transforms: linearity, time-product/differentiation, integration, time-shift, see for instance: Properties of Z-Transform:
$$ \mathcal{Z}(nx[n]) = -z\frac{d}{dz}X(z)\,,$$ and $$ \mathcal{Z}(x[n-n_0]) = z^{-n_0}X(z)\,.$$
Hint: if your case were a one-sided $\mathcal{Z}$-transform, one could have focused on derivation, and on the simple geometric series:
$$\sum_{n=0}^{+\infty}z^{-n} = \frac{z}{z-1}\,.$$
As of today, as pointed out by Matt and Robert, I could not make sense of the $$\sum_{n=-\infty}^{+\infty}z^{-n}\,,$$ and I can see but a zero radius of convergence. For what it's worth, I will anyway try to add a common trick to try to derive something. One can use a stabilization with some $0 < a <1$:
$$\sum_{n=-\infty}^{+\infty}a^{|n|}z^{-n}\,,$$ with the secret thought that, if $a$ is just close to one, we could get almost what we wanted. I may have made mistakes again, so far this yields the formal form:
$$ \frac{(1-a^2)z}{(z-a)(1-az)}\,.$$
This still have a singular ROC, but one can play with $\mathcal{Z}$-transform properties.
