Paul Sinclair's solution at https://math.stackexchange.com/a/2524119 gives a nice method for finding the limit of a sequence in which each term is the average of its preceding $k$ terms.
I wonder what the limit is if the sequence is a weighted average of its preceding $k$ terms, that is, what is $$\lim_{n\to\infty} a_{n}$$ for $$ a_{n} = \frac{w_{1}a_{n-1} + w_{2}a_{n-2} + \cdots + w_{k}a_{n-k}}{w_{1} + w_{2}+ \cdots +w_{k}} \;?$$
[tl;dr] $\;$ I don't think there is a nice closed form in the general case for arbitrary weights, short of solving a generic $(k-1)^{th}$ degree polynomial equation. The following, however, proves the existence of the limit, which was apparently taken as an assumption in the linked question.
A weighted average implies $w_j \ge 0$ with $w_1+w_2+\ldots+w_k \ne0$ so it can be assumed WLOG that $w_1+w_2+\ldots+w_k=1$ through the substitution $w_j \mapsto \frac{w_j}{w_1+w_2+\ldots+w_k}$.
The recurrence can then be written as $a_n=w_1a_{n-1}+w_2a_{n-2}+\ldots+w_ka_{n-k}$, which is a linear homogeneous recurrence with constant coefficients. The characteristic polynomial is $p(t)=t^k - w_1t^{k-1}-\ldots-w_{k-1}t-w_k=t^k q(\frac{1}{t})$ where $q(u)=w_ku^k+\ldots+w_1u-1$.
$p(t)=0 \iff q(u) = 0$ for $u= \frac{1}{t}$, or $u(w_ku^{k-1}+w_{k-1}u^{k-2}+\ldots+w_2 u+w_1)=1$. If $|u| \le 1$ then by the triangle inequality $|u(w_ku^{k-1}+w_{k-1}u^{k-2}+\ldots+w_2 u+w_1)|$ $\le |u|(w_k|u|^{k-1}+w_{k-1}|u|^{k-2}+\ldots+w_2 |u|+w_1)$ $\le w_k+w_{k-1}+\ldots+w_1=1$
Equality can only hold when $|u|=1$ and $u^{k-1}, u^{k-2},\ldots,1$ are positive scalar multiples of each other, so $u=1 \iff t=1$. It follows that all other roots of $q(u)$ are outside the unit circle $|u| \gt 1$ and therefore the roots of $p(t)$ are $t_1=1$ and $t_{2\,\ldots,k}$ with $|t_j| \lt 1$.
The general solution of the recurrence is $a_n = c_1 \cdot 1^n + c_2 \cdot t_2^n +\ldots + c_k \cdot t_k^n$. Since $|t_j|_{j=2,\ldots,k} \lt 1$ all but the first term vanish when $n \to \infty$, so $\lim_{n\to\infty} a_n = c_1$.
The constant $c_1$ can be determined from the initial conditions $(a_0, a_1, \ldots, a_{k-1})$:
$$ \begin{cases} \begin{align} a_0 &= c_1 + c_2 + \ldots + c_k \\ a_1 &= c_1 + c_2t_2+\ldots+c_kt_k \\ \dots \\ a_{k-1} &= c_1 + c_2t_2^{k-1}+\ldots+c_kt_k^{k-1} \\ \end{align} \end{cases} $$
Solving the linear system for $c_1$:
$$ c_1 = \frac{\begin{vmatrix} a_0 & 1 & \cdots & 1 \\ a_1 & t_2 & \cdots & t_k \\ \dots \\ a_{k-1} & t_2^{k-1} & \cdots & t_k^{k-1} \end{vmatrix}}{\begin{vmatrix} 1 & 1 & \cdots & 1 \\ 1 & t_2 & \cdots & t_k \\ \dots \\ 1 & t_2^{k-1} & \cdots & t_k^{k-1} \end{vmatrix}} $$
The numerator is not readily calculatable in the general case, since it's neither symmetric in $t_j$ (unless $a_n$ is a constant sequence, or the initial conditions are of some special form), nor otherwise easy to determine without knowing $t_j$ (as in the linked question about unweighted averages).
