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I want to find of the following series where $ F_0 = 1 $ , $ F_1 = 1 $ and $ F_n=F_{n-1}+F_{n -2} $ .

$$ S = \frac{F_0}{6}+\frac{F_1}{6^1} +\frac{F_2}{6^2}+\frac{F_3}{6^3}+\frac{F_4}{6^4} + \ldots$$

I have tried to proceed with the thought of finding a close formula for this problem but cant be able to derive formula . Can you please help me to find the formula for this problem ?

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    $\begingroup$ Hints: Binet's formula (for the Fibonaccis) and the sum formula for a geometric series. $\endgroup$ – Jyrki Lahtonen Sep 23 '13 at 6:39
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    $\begingroup$ equivalently (though it's somewhat simpler), use the generating function for $F_n$'s $\endgroup$ – user8268 Sep 23 '13 at 6:50
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    $\begingroup$ Actually what you "want to find" is the sum of the series, not the series (which you already have). $\endgroup$ – Did Sep 23 '13 at 7:02
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First, the series converges because $0 \leq F_n \leq 2^n$ by induction.

Then, you can compute $$ S = \sum_{n=0}^{+\infty} \frac{F_n}{6^n} = F_0 + \frac{F_1}{6} + \sum_{n=2}^{+\infty} \frac{F_n}{6^n} = F_0 + \frac{F_1}{6} + \sum_{n=2}^{+\infty} \frac{F_{n-1}}{6^n} + \sum_{n=2}^{+\infty} \frac{F_{n-2}}{6^n}. $$ Reordering should get you to $$ S = F_0 + \frac{F_1}{6} + \frac{S-F_0}{6} + \frac{S}{6^2}. $$ Then it's easy to find $S$.

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Using matrices I get this result:

Assume the matrix $M= \small \begin{bmatrix} 0&1\\1&1 \end{bmatrix}$
and the vector $R_1=[F_0,F_1]=[1,1]$,

then $R_1 \cdot M/6 = R_2 = [F_1/6,F_2/6]$,
and $R_1 \cdot (M/6)^2 = R_3 = [F_2/36,F_3/36]$
and so on with increasing exponents.

Now the infinite series $$ I+ (M/6) + (M/6)^2 + (M/6)^3 + ... = (I-M/6)^{-1} = \small \begin{bmatrix} 30/29&6/29\\6/29&36/29\end{bmatrix}$$ and let's denote the matrix as $W$ then your result is in

$$ R_1/6 \cdot W = [6/29, 7/29]$$ and because you asked for the series beginning with $F_0$ we use the first entry (at the position, where also $F_0$ is in $R_1$, so the result is $\qquad 6/29 \qquad$

(I hope, the indexes are correct, just have some trouble with my teeth...)

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