Evaluate the sum: $\sum\limits_{n=0}^{\infty} \frac1{F_{(2^n)}}$

Question

Evaluate the sum:

$$\sum_{n=0}^{\infty} \frac{1}{F_{(2^n)}}$$

where $F_{m}$ is the $m$-th term of the Fibonacci sequence. I need some support here. Thanks.

It is true that an answer to 105412 is an answer to the current question. However, there are answers to this question that would not qualify as answers to the older question, so I don't think this one qualifies as a duplicate. — Gerry Myerson, Jul 25 '12 at 01:50

score 6 · Accepted Answer · answered Jun 12 '12 at 19:01

6

As wikipedia claims the result follows from the identity $$ \sum\limits_{n=0}^N\frac{1}{F_{2^n}}=3-\frac{F_{2^N-1}}{F_{2^N}} $$ You can try to prove it by induction.

answered Jun 12 '12 at 19:01

Norbert

56,803

2

nice identity. I wonder what happens if one meets this problem in a competition because there you don't have access at wikipedia. :) Maybe there is a resonable approach for solving it. Thanks for information. – user 1591719 Jun 12 '12 at 19:17

score 5 · Answer 2 · answered Jun 12 '12 at 18:52

5

See http://oeis.org/A079585 and references there.

answered Jun 12 '12 at 18:52

Robert Israel

448,999

1

@Chris: The paper by Rabinowitz is available as a .ps file here. – Brian M. Scott Jun 12 '12 at 21:36
@Brian M. Scott: nice!!! Thanks for the paper. A good piece of information there! – user 1591719 Jun 12 '12 at 22:06

score 4 · Answer 3 · answered Jun 12 '12 at 18:56

4

Typing Sum[1/Fibonacci[2^n],{n,0,infty}] in http://www.wolframalpha.com gives,

$$\sum_{n=0}^\infty \frac{1}{F_{(2^n)}} = \left(\frac{1-\sqrt{5}}{2}\right)^3+\left(\frac{1+\sqrt{5}}{2}\right)^2 = 2.381966\dots$$

answered Jun 12 '12 at 18:56

Tito Piezas III

54,565

an interesting result. – user 1591719 Jun 12 '12 at 19:00
Your answer simplifies to $\frac{5-\sqrt5}2$. – Antonio DJC Jan 22 '18 at 06:26

Evaluate the sum: $\sum\limits_{n=0}^{\infty} \frac1{F_{(2^n)}}$

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