Convergence of a sequence - z-transform and Fourier transform When my book introduces the relationship between z-transform and Fourier transform, it starts from the following sequence ($x[n]$ generic sequence):
$$\widetilde{x}[n]=x[n] \space \left(\frac{1}{r}\right)^n$$
and it says that the above sequence convergs when $r>0$ and $n\rightarrow +\infty$. Why? If $0<r<1$, the sequence divergs.
Thank you very much.
 A: Teoria dei segnali, 1999 - Page 317
(Sorry for my rusty Italian, in case I miss something on the translation, ok?)
6.3 - Overview of the Z-Transform of a sequence
6.3.1 - Definition of the Z Transform and Region of Convergence
There is the sequence you provided:
$$\widetilde{x}[n]=x[n] \space \left(\frac{1}{r}\right)^n$$
Then he says that it is defined for $r>0$ and that it should quickly go to zero as $n \to \infty$. Then he uses as an example with Heaviside Step and then he states:
[...] However, the sequence $\widetilde{x}[n]$ defined as above is exponentially damped that has a transform if $r>1$ 
So, I also think there should be a better explanation about the values between $0$ and $1$, but he uses this "modified version" of the sequence just to introduce and define the Z-Transform from Fourier...
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EDIT:
Let us do the inverse analysis and try to crack it down a little more...
We know that the definition of the Z-transform is:
$$X(z)=\mathcal{Z}\{x[n]\}=\sum_{n=-\infty}^{+\infty}x[n]z^{-n}$$
But, we know thta $z\in \mathbb{C}$, so that $z=re^{j\omega}$ and $r=|z|$. That way, we can rewrite:
$$
\begin{align}
X(z)&=\sum_{n=-\infty}^{+\infty}x[n](re^{j\omega})^{-n}\\
&=\sum_{n=-\infty}^{+\infty}x[n]r^{-n}e^{-j\omega n}\\
&=\sum_{n=-\infty}^{+\infty}x[n]\left(\frac{1}{r}\right)^{n}e^{-j\omega n}\\
&=\sum_{n=-\infty}^{+\infty}\widetilde{x}[n]e^{-j\omega n}=DTFT\{\widetilde{x}[n]\}
\end{align}
$$
So, we get to the relation:
$$\mathcal{Z}\{x[n]\}=DTFT\{\widetilde{x}[n]\}$$
With special case when $r=1$, having:
$$\mathcal{Z}\{x[n]\}=DTFT\{x[n]\}$$
This is the Discrete-Time equivalent analogy to Continuous-Time Fourier Transform incapability of dealing witrh unlimited energy signals, which causes the Laplace Transform to take place:
$$X(s)=\mathcal{L}\{x(t)\}=\int_{0}^{+\infty}x(t)e^{-st}dt$$
We know that $s \in \mathbb{C}$ and $s=\sigma + j\omega$,giving us:
$$
\begin{align}
X(s)&=\int_{-\infty}^{+\infty}x(t)e^{-(\sigma + j\omega)t}dt\\
&=\int_{-\infty}^{+\infty}x(t)e^{-\sigma t}e^{-j\omega t}dt\\
&=\int_{-\infty}^{+\infty}\widetilde{x}(t)e^{-j\omega t}dt=\mathcal{F}\{\widetilde{x}(t)\}
\end{align}
$$
So, for a given continuous-time, exponentially damped signal
$$\widetilde{x}(t)=x(t)e^{-\sigma t}$$
We have the relation:
$$\mathcal{L}\{x(t)\}=\mathcal{F}\{\widetilde{x}(t)\}$$
With special case when $\sigma=0$, giving:
$$\mathcal{L}\{x(t)\}=\mathcal{F}\{x(t)\}$$
Which is widely used to trace Bode Diagrams of systems described in Frequency Domain.
One more thing: I believe the book just defined $r>0$ because $r\stackrel{\Delta}=|z|$. Of course it does not imply that is convergent for all values of $r$ in that interval, after all, a divergent series $x[n]$ has a Z-transform, with pole(s) outside the unit circle. The convergence of $\widetilde{x}[n]$ is useful on the analysis of the $ROC$, which I can add to a further Edit if you need/want so.
