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In connection with a problem I'm solving, I seem to be getting the series $$S = 4 \cdot \frac{F_1}{4}+5 \cdot \frac{F_2}{8}+6 \cdot \frac{F_3}{16}+7 \cdot \frac{F_4}{32}+ \cdots$$ where $F_i$ are the Fibonacci numbers. How does one solve this?

I can see here that the Fibonacci sequence is related to a geometric sequence, but I'm not sure how to incorporate that.

Jens
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    Well, I'd start by computing the generating function $F(x)=\sum F_nx^n$. – lulu Oct 16 '17 at 17:55
  • @lulu: It appears I need to read up on generating functions. Could you provide a bit more detail? Is $S = F(x)$? How do I incorporate the arithmetic part into that expression? – Jens Oct 16 '17 at 18:13
  • I'm writing up more details. I keep making algebraic blunders so it might take a few minutes. – lulu Oct 16 '17 at 18:14

2 Answers2

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Let $$F(x) =\sum _{i=0}^{\infty}F_nx^n$$ Then by Infinite Series: Fibonacci/ $2^n$ we have

$$F(x) = {x\over 1-x-x^2}$$ we are interested in $F'(1/2)$ so ...

nonuser
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Consider the generating function $$F(x)=\sum_{n=1}^{\infty}F_nx^n$$ If we index the Fibonacci numbers as $F_1=1=F_2$ then this can be computed as $$F(x)=\frac x{1-(x+x^2)}$$

See, e.g., this question

We now just have to adapt this to suit your series.

First remark that $$G(x)=x^3F(x)=\frac {x^4}{1-(x+x^2)}=\sum_{n=1}^{\infty}F_nx^{n+3}$$ It follows that $$\frac d{dx}G(x)=3x^2F(x)+x^3F'(x)=\sum_{n=1}^{\infty}(n+3)F_nx^{n+2}$$

Taking $x=\frac 12$ yields the sum $4\frac {F_1}8+5 \frac {F_2}{16}+\cdots$ which is eaxactly half of your sum. Thus the answer you seek is $$2\times \left(3\times \left(\frac 12\right)^2\times F\left(\frac 12\right)+\left(\frac 12\right)^3F'\left(\frac 12\right)\right)$$

This is (relatively) easy to compute and yields $$\boxed 8$$

Worth remarking: This agrees with numerical computation (summing the first hundred terms yields $7.999999936$).

lulu
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  • Thank you! I'm going to have to read this several times to understand what is going on, but the fact that you arrived at the answer my simulations suggested, gives great confidence. – Jens Oct 16 '17 at 18:24
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    You're welcome. Just to say, the numerical confidence is essential. I had a small, but hard to spot, error that kept giving me $14$. Without the numerical check I'd have been inclined to believe it. For that reason, I always like to have the numerical value in hand. – lulu Oct 16 '17 at 18:27
  • If you are interested in the problem I was working on, it is here. Thank you again. And let me know what you think about my answer there. – Jens Oct 16 '17 at 19:48