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.
 A: Outline: This follows the spirit of Thomas Andrew's solution to the OP's other question to find the value of the series.  Convergence is then proven by showing $S_n/p^n<1$ for sufficiently large p. 

Finding the Value with Generating Functions:
Define $S_n$ for a fixed $k$ as follows:
$$S_n = \begin{cases} 0 & n \leq 0 \\
1 & n = 1 \\
\sum_{j=1}^k S_{n-j} & n > 1
\end{cases}$$
Let $\mathcal{S}(z) = \sum_{n\geq 0}S_nz^n$.  I assert without proof (I'll seek a source paper, rather than derive it myself) that:
$$\mathcal{S}(z) = \frac{z}{1 - z - z^2 -\dots - z^k}$$
It follows:
\begin{align*}
\mathcal{S}(z) &= \frac{z}{1 - z - z^2 -\dots - z^k} \\
&= \frac{z}{2 - \sum_{j=0}^k z^j} \\ 
&= \frac{z}{2 - \frac{z^{k+1}-1}{z-1}} \\ 
&= \frac{z(z-1)}{2z -z^{k+1}-1} \\ 
&= \frac{z(z-1)}{(2-z^k)z-1}
\end{align*}
The value we seek is $\mathcal{S}(1/p)$:
\begin{align*}
\mathcal{S}(1/p) &= \frac{(1/p)\left(\frac{1}{p}-1\right)}{\left(2-\frac{1}{p^k}\right)\frac{1}{p}-1} \\
&= \frac{\frac{1-p}{p^2}}{\left(\frac{2p^k-1}{p^{k+1}}\right)-1} \\
&= \frac{1-p}{\frac{p^2}{p^{k+1}}\left(2p^k-1 -p^{k+1}\right)} \\
&= \frac{(1-p)p^{k-1}}{(2-p)p^k -1}
\end{align*}
This is as desired.

Proving Convergence:
However, this only provides the value if the series converges; that proof is entirely separate. 
It is generally known that $S_n \in \mathcal O( r^n )$, where $r$ is the largest real root of the equation $2-\sum_{j=0}^k z^j = 0$.  To justify this, consider the last identity in this Wiki article and consider its asymptotic behavior.  (I don't like the source of that, but I'll find a better one later).
Thus, so long as $r < p$, the series $\mathcal{S}(1/p)$ is bounded above by a convergent geometric series $\sum_{n\geq 0} \left(\frac{r}{p}\right)^n$.
Fortunately, based on Lemma 2.7 of this arxiv paper, we know that the largest real root of $2-\sum_{j=0}^kz^j = 0$ is strictly bounded between $1$ and $2$.  Thus, so long as $p \geq 2$, we have convergence.
A: 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}
A: That's because
$\sum_{n=0}^{\infty} F_nx^n
=\frac1{1-x-x^2}
$.
Putting
$x = \frac1{10^k}$
gives
$\sum_{n=0}^{\infty} \frac{F_n}{10^{kn}}
=\frac1{1-10^{-k}-10^{-2k}}
=\frac{10^{2k}}{10^{2k}-10^{k}-1}
$.
For the others,
the generating function is
$\sum_{n=0}^{\infty} G_n x^n
=\frac1{1-x-x^2-...-x^m}
$
where
$m=2$ for Fib,
$m=3$ for Trib,
and $m=4$ for Tetra.
For each of these,
$\sum_{n=0}^{\infty} \frac{G_n}{ 10^n}
=\frac1{1-\frac1{10}-\frac1{100}-...-\frac1{10^{m}}}
=\frac{10^m}{10^m-10^{m-1}-...-1}
=\frac{10^m}{8...(m-1 \ 8s)9}
$.
