Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
Hints: Binet's formula (for the Fibonaccis) and the sum formula for a geometric series. –  Jyrki Lahtonen Sep 23 '13 at 6:39
2  
equivalently (though it's somewhat simpler), use the generating function for $F_n$'s –  user8268 Sep 23 '13 at 6:50
1  
Actually what you "want to find" is the sum of the series, not the series (which you already have). –  Did Sep 23 '13 at 7:02

2 Answers 2

up vote 8 down vote accepted

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 $S$.

share|improve this answer

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...)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.