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.

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}
$.
