12

let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$

Find this value

$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$

My try: I know this

$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$$

so

$$\dfrac{1}{F_{2^k}}=\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$ so we only find this sum $$I=\sum_{k=0}^{\infty}\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$

But I can't.Thank you

math110
  • 93,304
  • 1
    There is a generalization which has this as a special case. See http://math.stackexchange.com/questions/157820 – Tito Piezas III Nov 24 '13 at 06:52
  • @Tito: Oh, I feel so deflated for my proof now. I was so proud. I feel worse so, because I not only upvoted your question - I favorited it too. I suppose I should have been suspect when I came up with my answer so relatively quickly. – davidlowryduda Nov 24 '13 at 06:55
  • @mixedmath: Sorry. At least you knew how to prove it. :) – Tito Piezas III Nov 24 '13 at 06:58

2 Answers2

12

In fact, this is summable. This surprises me, because of results like those in this question at MO, which in short discusses how we don't know if the sum of the reciprocals of all the Fibonacci numbers is transcendental, and other related unknown results.

This question boils down to the following:

Claim: $\displaystyle \frac{1}{F_1} + \frac{1}{F_2} + \frac{1}{F_4} + \ldots + \frac{1}{F_{2^n}} = 3 - \frac{F_{2^n-1}}{F_{2^n}}$ for $n \geq 2$.

Proof: Induct, and use $F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$ as you mentioned. $\diamondsuit$

Then take the limit as $n \to \infty$ to get that

$$\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}} = \frac{7-\sqrt 5}{2}.$$

Community
  • 1
davidlowryduda
  • 91,687
12

let $$a=\dfrac{1+\sqrt{5}}{2}\Longrightarrow \dfrac{\sqrt{5}-1}{2}=a^{-1}$$ so $$I=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}$$ so let $$a^{2^n}=x$$ then $$\dfrac{1}{a^{2^n}-a^{-2^n}}=\dfrac{x}{x^2-1}=\dfrac{1}{x-1}-\dfrac{1}{x^2-1}=\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}$$ so \begin{align*}I&=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}=\sqrt{5}\sum_{n=0}^{\infty}\left(\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\ &=\lim_{n\to\infty}\left(\dfrac{\sqrt{5}}{a-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\ &=\dfrac{\sqrt{5}}{a-1}\\ &=\dfrac{7-\sqrt{5}}{2} \end{align*}

math110
  • 93,304