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Compute \(\displaystyle \sum_{n = 0}^\infty \frac{F_n}{4^n}\)

 

Note: F_n denotes the nth Fibonacci number.

 Jun 16, 2020
 #1
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sumfor(n, 0, 1000, (fibon(n) / 4^n) = 4 / 11

 Jun 16, 2020
 #2
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More generally, we consider the sum \(S = \displaystyle\sum_{n = 0}^\infty F_n x^n\)\(|x| < \dfrac{\sqrt 5 - 1}2\).

Here, the case is x = 1/4, but we will consider that later.

 

By Binet's formula, \(F_n = \dfrac{(1 + \sqrt 5)^n - (1 - \sqrt 5)^n}{2^n \sqrt 5}\).

 

Substituting this into the sum and denote \(\dfrac{1 + \sqrt 5}2 \) as \(\varphi\),

\(S = \displaystyle\sum_{n = 0}^\infty F_n x^n = \dfrac1{\sqrt 5} \sum_{n = 0}^\infty \left((\varphi x)^n - \left(\dfrac{-x}{\varphi}\right)^n\right)\)

 

This is the sum of G.S., and can be calculated by the formula \(\displaystyle\sum_{n = 0}^\infty r^n = \dfrac{1}{1 - r}\).

 

Therefore \(S = \dfrac1{\sqrt{5}} \left(\dfrac1{1 - x\varphi} - \dfrac1{1 + \dfrac{x}{\varphi}}\right)\).

 

Now, substituting x = 1/4,

 

\(S = \dfrac1{\sqrt 5} \left(\dfrac1{1 - \dfrac{\varphi}4} - \dfrac1{1 + \dfrac1{4\varphi}}\right)\)

 

Upon simplification, we have 

 

\(S = \dfrac1{\sqrt 5} \left(\dfrac{4\sqrt 5}{11}\right) = \boxed{\dfrac 4{11}}\)

 

The computation I did with Wolfram: https://www.wolframalpha.com/input/?i=1%2F%281+-+%28golden+ratio%29%2F4%29+-+1%2F%281+%2B+1%2F%284%28golden+ratio%29%29%29

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 Jun 17, 2020
edited by MaxWong  Jun 17, 2020

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