Fibonacci numbers and powers of two

Calculus Level 3

k = 0 1 F 2 k \large \sum_{k=0}^\infty \frac{1}{F_{2^k}}

Evaluate the sum above, where F n F_n is the n n th Fibonacci number . Round your answer to the second decimal place.


The answer is 2.38.

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2 solutions

Mark Hennings
Aug 16, 2018

Using Binet's Formula for the Fibonacci numbers, F n = 1 5 [ ϕ n ( ϕ 1 ) n ] n 0 F_n \; = \; \frac{1}{\sqrt{5}}\big[\phi^n - (-\phi^{-1})^n\big] \hspace{2cm} n \ge 0 where ϕ = 1 2 ( 1 + 5 ) \phi = \tfrac12(1 + \sqrt{5}) is the golden ratio, we see that S = n = 0 F 2 n = n = 0 5 ϕ 2 n ( ϕ 1 ) 2 n = n = 0 5 ϕ 2 n ( 1 ( ϕ 2 ) 2 n ) 1 = 5 ϕ ( 1 + ϕ 2 ) + n = 1 5 ϕ 2 n ( 1 ϕ 2 n + 1 ) 1 = 5 ϕ ( 1 + ϕ 2 ) + n = 1 m = 0 5 ϕ ( 2 m + 1 ) 2 n \begin{aligned} S \; = \; \sum_{n=0}^\infty F_{2^n} & = \; \sum_{n=0}^\infty \frac{\sqrt{5}}{\phi^{2^n} - (-\phi^{-1})^{2^n}} \; = \; \sum_{n=0}^\infty \frac{\sqrt{5}}{\phi^{2^n}}\left(1 - (-\phi^{-2})^{2^n}\right)^{-1} \\ & = \; \frac{\sqrt{5}}{\phi(1 + \phi^{-2})} + \sum_{n=1}^\infty \frac{\sqrt{5}}{\phi^{2^n}}\left(1 - \phi^{-2^{n+1}}\right)^{-1} \; = \; \frac{\sqrt{5}}{\phi(1 + \phi^{-2})} + \sum_{n=1}^\infty \sum_{m=0}^\infty \frac{\sqrt{5}}{\phi^{(2m+1)2^n}} \\ \end{aligned} Now the numbers ( 2 m + 1 ) 2 n (2m+1)2^n for m 0 m \ge 0 and n 1 n \ge 1 simply run through all the even positive integers. Thus we deduce that S = 5 ϕ ( 1 + ϕ 2 ) + n = 1 5 ϕ 2 n = 5 ϕ ( 1 + ϕ 2 ) + 5 ϕ 2 1 ϕ 2 = 5 ϕ 1 + ϕ 2 + 5 ϕ 2 1 = 5 ( 3 ϕ 1 ) ϕ + 2 = 1 2 ( 7 5 ) S \; = \; \frac{\sqrt{5}}{\phi(1 + \phi^{-2})} + \sum_{n=1}^\infty \frac{\sqrt{5}}{\phi^{2n}} \; = \; \frac{\sqrt{5}}{\phi(1 + \phi^{-2})} + \frac{\sqrt{5}\phi^{-2}}{1 - \phi^{-2}} \; = \; \frac{\sqrt{5}\phi}{1 + \phi^2} + \frac{\sqrt{5}}{\phi^2-1} \; = \; \frac{\sqrt{5}(3\phi-1)}{\phi + 2} \; = \; \boxed{\tfrac12(7-\sqrt{5})}

@Jeremy Galvagni

Nice. I am not well versed with the \sum stuff, but it looks cool.

Jeremy Galvagni - 2 years, 9 months ago
Jeremy Galvagni
Aug 14, 2018

This converges extremely quickly. F 2 6 > 1 0 13 F_{2^{6}} > 10^{13} so the first 7 terms give a sum 2.381966011 2.381966011 on my calculator which is accurate to all displayed digits.

Out of interest, I put this into Inverse Symbolic Calculator which had the single hit 7 5 2 \frac{7-\sqrt{5}}{2} which seems reasonable as the exact solution.

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