$$\omega=(\omega_1, \omega_2,...)\in \{0, 1, 2,...,N-1\}^\mathbb N$$
How to show that the limit$$\displaystyle\nu(\omega)=\lim_{n\to\infty}\frac{\omega_1}{N^n}+\frac{\omega_2}{N^{n-1}}+...+\frac{\omega_n}{N}$$ exist?
it can be represented as a $N$-adic number in $[0, 1]$, but the sequence $\frac{\omega_1}{N^n}+\frac{\omega_2}{N^{n-1}}+...+\frac{\omega_n}{N}$ is not monotonic. I can only show that it has limit if $\omega_i=\omega\in\{0,1,...,N-1\}$ for $i\ge$ some number.
I am reading Analysis and Probability: Wavelets, Signals, Fractals, on page 92, it states that this sequence is non-increasing, is it right?
First of, the formulation is unfortunate. In the remainder class ring $\Bbb Z/N\Bbb Z$ there is no division by $N$, since $N$ is (in) the class of zero.
Your second formulation is the correct one, for a N-adic representation of a real number, you want $ω_i∈\{0,1,...,N−1\} \subset \Bbb Z\subset\Bbb R$.
And then you just use that $0\le ω_i\le N-1$ to get a convergent geometric series as majorant.
Sorry, I did initially not see that there is an index reversal in the terms of the sum. So the geometric majorant only shows that this sequence is bounded. Write your series as quotient of two divergent terms $$ \frac{ω_1N+ω_2N^2+...+ω_nN^n}{N^{n+1}} $$ but Cesaro-Stolz can only be applied if $$ \frac{ω_nN^n}{N^{n+1}-N^n}=\frac{ω_n}{N-1} $$ converges.
Give the sequence a name, $a_n=ω_1N^{-n}+ω_2N^{1-n}+...+ω_nN^{N-1}$, then $$ a_{n+1}=\frac{a_n+ω_{n+1}}{N}=\tfrac1Na_n+(1-\tfrac1N)\frac{ω_{n+1}}{N-1} $$ so $a_{n+1}\le a_n$ if and only if $ω_{n+1}\le (N-1)a_n$ which would require that $a_n\ge 1$ since $ω_{n+1}=N-1$ is possible. But since $a_n=0$ is also possible, the statement "non-increasing" is wrong. The best one can say is still that the sequence of the $a_n$ is bounded.