Reduced Fibonacci Series

Algebra Level 4

It is certain that the infinite sum of Fibonacci numbers is infinite. Consider this sum:

∑ n = 1 ∞ F n 2 n + 1 . \sum _{ n=1 }^{ \infty }{ \frac { { F }_{ n } }{ { 2 }^{ n+1 } } } .

If F 1 = 1 , F 2 = 1 , F 3 = 2 , . . . e t c {F}_{1} = 1, {F}_{2} = 1, {F}_{3} = 2, ... etc , what is the value of the above sum?


The answer is 1.

Steven Zheng by Steven Zheng , 26, China

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1 solution

Tijmen Veltman
Aug 21, 2014

Let S : = ∑ n = 1 ∞ F n 2 n + 1 S:=\sum_{n=1}^\infty \frac{F_n}{2^{n+1}} . Seeing as F n + 2 = F n + 1 + F n F_{n+2}=F_{n+1}+F_n for all n ∈ N n\in\mathbb{N} , we have:

S + 2 S = ∑ n = 1 ∞ F n 2 n + 1 + ∑ n = 1 ∞ F n 2 n = ∑ n = 1 ∞ F n 2 n + 1 + F 1 2 + ∑ n = 1 ∞ F n + 1 2 n + 1 = F 1 2 + ∑ n = 1 ∞ F n + 2 2 n + 1 = F 1 2 − F 1 − F 2 2 + ∑ n = 1 ∞ F n 2 n − 1 = − 1 + 4 S S+2S = \sum_{n=1}^\infty \frac{F_n}{2^{n+1}} + \sum_{n=1}^\infty \frac{F_n}{2^{n}} \\ = \sum_{n=1}^\infty \frac{F_n}{2^{n+1}} + \frac{F_1}2 + \sum_{n=1}^\infty \frac{F_{n+1}}{2^{n+1}} \\ = \frac{F_1}2 + \sum_{n=1}^\infty \frac{F_{n+2}}{2^{n+1}} \\ = \frac{F_1}2 - F_1-\frac{F_2}2 + \sum_{n=1}^\infty \frac{F_n}{2^{n-1}} \\ = -1 + 4S .

Hence 3 S = 4 S − 1 3S=4S-1 , giving us S = 1 \boxed{S=1} .

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