Let $F_0 = 0, F_1 = 1,$ and $F_n = F_{n-1}+F_{n-2}$. Find the value of the infinite sum $$\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{27}+\cdots+\dfrac{F_n}{3^n}+\cdots.$$

This sort of looks like an arithmetico-geometric series but except for the fact that the fiboncci sequence is not arithmetic. I couldn't think of a way to continue.

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    $\begingroup$ Hint: The generating function for the Fibonacci numbers is $\sum F_nx^n=\frac 1{1-(x+x^2)}$. $\endgroup$ – lulu Dec 12 '15 at 21:45
  • $\begingroup$ Do you know the generating function for the Fibonacci sequence? $\endgroup$ – Travis Dec 12 '15 at 21:45
  • $\begingroup$ Does that generating function work for infinite sums only? $\endgroup$ – Puzzled417 Dec 12 '15 at 21:46
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    $\begingroup$ Hint: what is the closed formula of Fn? $\endgroup$ – Martín Vacas Vignolo Dec 12 '15 at 21:46
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    $\begingroup$ @Puzzled417 there's a version for finite sums. Let $f_n(x)$ be the terms of the generating function up to $x^n$, then look at $f_n(x)-xf_n(x)-x^2f_n(x)$ and use the recursion. Most terms go away, and you can just gather up the rest. $\endgroup$ – lulu Dec 12 '15 at 21:56

Let $s_n=F_n/3^n$, we have: $$ s_n={1\over3}s_{n-1}+{1\over9}s_{n-2}, $$ that is:

$$ \sum_{k=3}^\infty s_n={1\over3}\sum_{k=2}^\infty s_{n}+{1\over9}\sum_{k=1}^\infty s_{n}. $$ If $S=\sum_{k=1}^\infty s_{n}$ is finite (which is true, because $s_n<(2/3)^n$), it follows that: $$ S-{1\over3}-{1\over9}={1\over3}\left(S-{1\over3}\right)+{1\over9}S, $$ whence $S=3/5$.

  • $\begingroup$ I don't understand you last line. $\endgroup$ – Puzzled417 Dec 12 '15 at 22:16
  • $\begingroup$ $\sum_{k=3}^\infty s_n=\sum_{k=1}^\infty s_n-s_1-s_2=S-1/3-1/9$ and $\sum_{k=2}^\infty s_n=\sum_{k=1}^\infty s_n-s_1=S-1/3$. $\endgroup$ – Aretino Dec 12 '15 at 22:19
  • $\begingroup$ What about $s_0$? You seem to be shifting the index of the series around. $\endgroup$ – Puzzled417 Dec 12 '15 at 22:20
  • $\begingroup$ But your sum starts with $s_1$! And anyway $s_0=0$. $\endgroup$ – Aretino Dec 12 '15 at 22:22
  • $\begingroup$ Sorry. You are right. $\endgroup$ – Puzzled417 Dec 12 '15 at 22:23

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