Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$

Question

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.

Answer 1

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.

Answer 2

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}

Answer 3

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} $.