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.

  • 2
    $\begingroup$ 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? $\endgroup$ Dec 12 '15 at 21:45
  • $\begingroup$ Does that generating function work for infinite sums only? $\endgroup$
    – Puzzled417
    Dec 12 '15 at 21:46
  • 1
    $\begingroup$ Hint: what is the closed formula of Fn? $\endgroup$ Dec 12 '15 at 21:46
  • 1
    $\begingroup$ @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 $\sum_{k=2}^\infty s_n=\sum_{k=1}^\infty s_n-s_1=S-1/3$. $\endgroup$ 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 $s_0=0$. $\endgroup$ Dec 12 '15 at 22:22
  • $\begingroup$ Sorry. You are right. $\endgroup$
    – Puzzled417
    Dec 12 '15 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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