Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
$$\sum_{n=0}^{\infty}\frac{T_n}{10^n} = \frac{100}{889}$$
$$\sum_{n=0}^{\infty}\frac{U_n}{10^n} = \frac{1000}{8889}$$
or, in general,
$$\sum_{n=0}^{\infty}\frac{S_n}{p^n} = \frac{(1-p)p^{k-1}}{(2-p)p^k-1}$$
where the above were just the cases $k=2,3,4$, and $p=10$?
P.S. Related post.

The difference equations given by the suggested series are:
\begin{align}
F_{n+2} &= F_{n+1} + F_{n} \\
T_{n+3} &= T_{n+2} + T_{n+1} + T_{n} \\ \tag{1}
U_{n+4} &= U_{n+3} + U_{n+2} + U_{n+1} + U_{n}
\end{align}
and so on. In general they take on the form
\begin{align}\tag{2}
\phi_{n+m} = \sum_{k=0}^{m-1} \phi_{n+m-k-1},
\end{align}
where $\phi_{0}, \phi_{1}, \phi_{2}, \cdots $ are the initial values.
By considering the generating function defined by
\begin{align}
f_{m}(t) = \sum_{n=0}^{\infty} \phi_{n+m} \, t^{n}
\end{align}
then it is readily found that
\begin{align}
f_{m}(t) &= \frac{1}{ 2 - \sum_{k=0}^{m} t^{k}} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \\
&= \frac{1 - t}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \tag{3}
\end{align}
if $t \to 1/t$ then
\begin{align}
f_{m}(t) &= \frac{t^{m} (t - 1)}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, \frac{1}{t^{k}} \right]
\end{align}
When $t = 10$ this reduces to
\begin{align}\tag{4}
f_{m}\left(\frac{1}{10}\right) = \frac{9}{(10)^{m+1}- 2 \, (10)^{m} + 1} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, (10)^{m-k} \right]
\end{align}
As an example let $m=3$, which corresponds to the Tribonacci series, to obtain
\begin{align}
f_{3}\left(\frac{1}{10}\right) &= \sum_{n=0}^{\infty} \frac{T_{n}}{(10)^{n}} 
= \frac{9 \, (10)^{3}}{10^{4} - 2 \cdot 10^{3} + 1} \cdot \left(\frac{1}{10}\right) = \frac{100}{889}.
\end{align}
