# Find the sum of the infinite Fibonacci sequence

Problem

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.

• Hint: The generating function for the Fibonacci numbers is $\sum F_nx^n=\frac 1{1-(x+x^2)}$.
– lulu
Dec 12 '15 at 21:45
• Do you know the generating function for the Fibonacci sequence? Dec 12 '15 at 21:45
• Does that generating function work for infinite sums only? Dec 12 '15 at 21:46
• Hint: what is the closed formula of Fn? Dec 12 '15 at 21:46
• @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.
– 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 $S=3/5$.
• $\sum_{k=3}^\infty s_n=\sum_{k=1}^\infty s_n-s_1-s_2=S-1/3-1/9$ and $\sum_{k=2}^\infty s_n=\sum_{k=1}^\infty s_n-s_1=S-1/3$. Dec 12 '15 at 22:19
• What about $s_0$? You seem to be shifting the index of the series around. Dec 12 '15 at 22:20
• But your sum starts with $s_1$! And anyway $s_0=0$. Dec 12 '15 at 22:22