Enlarging set where "weighted" MacLaurin series of $\frac{1}{1 - x}$ equals $\frac{1}{1 - x}$ Is it possible to select real values $a_{n, k}$ so that
$$f(x) =\lim_{n \to \infty}\sum_{k = 0}^{n - 1} a_{n, k} x^k = \frac{1}{1 - x} $$ for all $x \in \mathbb{R} \setminus \{1\}$ ?
Failing examples:


*

*$a_{n, k} = 1$ for all $n, k \in \mathbb{N}_0$ and $\lvert x \rvert < 1$ 


$$f(x) = \lim_{n \to \infty} \sum_{k = 0}^{n - 1} x^k = \frac{1}{1 - x}$$


*$a_{n, k}  = 1 - \frac{k}{n}$ for all $n, k \in \mathbb{N}_0$ and $ x\in [-1, 1)$ (One extra point!) 


$$\begin{aligned} f(x) &=\lim_{n \to \infty}\sum_{k = 0}^{n - 1} \left( 1 - \frac{k}{n}\right) x^k \\ &=\lim_{n \to \infty}\sum_{k = 0}^{n - 1} \left( 1 - \frac{k}{n}\right) x^k \\ 
&= \lim_{n \to \infty} \frac{x (x^n-1)}{n (x-1)^2} + \frac{1}{1-x} \\ &= \frac{1}{1 - x}\end{aligned}$$
 A: By the Stone-Weierstraß theorem, we can uniformly approximate every continuous function on a compact subset of $\mathbb{R}$ by a sequence of polynomials.
Thus if we exhaust $\mathbb{R}\setminus \{1\}$ by a sequence of compact sets, say $K_m = \{ x \in \mathbb{R} : 2^{-m} \leqslant \lvert x-1\rvert \leqslant 2^m\}$, and for each $m$ pick a polynomial $p_m$ such that
$$\sup \left\{ \biggl\lvert p_m(x) - \frac{1}{1-x}\biggr\rvert : x \in K_m\right\} \leqslant 2^{-m}$$
and $\deg p_{m+1} > \deg p_m$ for all $m$, we obtain a sequence of polynomials converging pointwise - even locally uniformly - to $\frac{1}{1-x}$ on $\mathbb{R}\setminus \{1\}$.
Letting $a_{n,k}$ the coefficients of $p_m$ for $\deg p_m < n \leqslant \deg p_{m+1}$, we then have
$$\lim_{n\to\infty} \sum_{k = 0}^{n-1} a_{n,k} x^k = \frac{1}{1-x}$$
for all $x \in \mathbb{R}\setminus \{1\}$ as desired.
However, I don't think one could call such a sequence a "weighted MacLaurin series" in any reasonable way. The coefficients of the $p_m$ have most likely very little to do with the coefficients of the Taylor/MacLaurin series of $\frac{1}{1-x}$.
A: In order that our power sums match the Taylor series of $\frac{1}{1-x}$ in a neighbourhood of the origin we must have $\lim_{n\to +\infty}a_{n,k}=1$ for every $k$, so we cannot have convergence outside the unit disk.
