How to take the limit of a set of functions $\newcommand{\range}{\operatorname{Range}}$I have a function with an infinite number of variables which can be produced as the limit of a set of finite functions with increasingly many variables.
$x_i \in \mathbb{N}$ (does not include $0$)
$T_1(x_1) = \frac{2^{x_1}}{3^1}-\frac{3^0}{3^1}$
$T_2(x_1,x_2) = \frac{2^{x_1+x_2}}{3^2}-\frac{3^0\times 2^{x_1}+3^1}{3^2}$
$T_3(x_1,x_2,x_3) = \frac{2^{x_1+x_2+x_3}}{3^3}-\frac{3^0\times 2^{x_1+x_2}+3^1\times2^{x_1}+3^2}{3^3}$
$$T_d(x_1, \dots, x_d)= \frac{2^{x_1+\dots+x_d}}{3^d}-\frac{1}{3^d}\sum_{a=1}^{d-1} 3^{d-a-1}\times2^{x_0+\dots+x_a}$$
$$T=\lim_{d\to\infty} T_d(x_1,\dots,x_d)$$
I'd like to know the range of T. I suspect it is:
$$\range(T)=\left\{\frac{2k-1}{3^{n-1}}:k,n \in \mathbb{N}\right\}$$ But I have no idea how to prove this. As I have no experience dealing with functions formulated in this manner. Any help / advice is greatly appreciated.


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*It may be worth noting that if all $x_i = 2$ with $i>d$ then $T(x_1, \dots, x_d) = T(x_1,\dots,x_d,2,2,2,2,2,\cdots)$

 A: First calculate the range of $T_d$ for all $d$.$\newcommand{\range}{\operatorname{Range}}$
Then, just show that for all $d$, $\range(T_d) \subset \range(T)$ -- reading the comments it seems like you have already done that.
Then you have that $\bigcup_{d=1}^{\infty} \range(T_d) \subset \range(T) $. I assume that $\bigcup_{d=1}^{\infty} \range(T_d) =\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\}$ although admittedly I haven't put much thought into what $\range(T_d)$ must be.
Thus, to show that $\range(T)=\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\} $, assuming that $\bigcup_{d=1}^{\infty} \range(T_d) =\left\{ \frac{2k-1}{3^{n-1}}:k,n\in \mathbb{N} \right\}$, all that remains to show is the opposite inclusion, i.e. $$\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d).  $$ So let $y \in \range(T)$, show that there must exist some $d$ such that $y=T(x_1,\dots, x_d, 2, 2, \dots)=T_d(x_1, \dots, x_d)$, conclude that $y \in \range(T_d) \subset \bigcup_{d=1}^{\infty} \range(T_d)$ and thus that $\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d)$.
Since $\range(T) \subset \bigcup_{d=1}^{\infty} \range(T_d)$ and $\bigcup_{d=1}^{\infty} \range(T_d) \subset \range(T)$, one would have $\range(T)=\bigcup_{d=1}^{\infty} \range(T_d)$ and you would be done.
