Fibonacci Product

Algebra Level 5

n = 2 ( 1 + 2 F 2 n 1 ) \large \prod_{n=2}^\infty \left(1 + \dfrac2{F_{2n} - 1} \right)

Let F n F_n denote the n th n^\text{th} Fibonacci number , where F 0 = 0 , F 1 = 1 F_0 = 0, F_1 = 1 and F n = F n 1 + F n 2 F_n = F_{n-1} + F_{n-2} for n = 2 , 3 , 4 , n=2,3,4,\ldots .

Compute the infinite product above.


The answer is 3.

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1 solution

Joel Tan
Oct 28, 2016

After some tedious exploration:

Denote k 0 = 0 k_0=0 and k i = k i 1 + F 2 i k_{i} = k_{i-1} + F_{2i} for all positive integers i i , (i.e., the sequence is 1, 4, 12, 33, ...) then

1 + 2 F 2 n 1 = ( k n 1 + 2 k n 1 ) ( k n k n + 2 ) 1+\frac{2}{F_{2n}-1} = (\frac{k_{n-1}+2}{k_{n-1}})(\frac{k_{n}}{k_{n}+2}) .

Now if we express each term 1 + 2 F 2 n 1 1+\frac{2}{F_{2n}-1} in the form above, it telescopes (the terms cancel out) leaving only the first term, and since k n k n + 2 \frac{k_{n}}{k_{n}+2} tends to 1 as n n becomes huge, the product will converge to k 1 + 2 k 1 = 3 \frac{k_{1}+2}{k_{1}}=3 .

(As for why it works, consider the expansion of F n = 1 5 ( ( 1 + 5 2 ) n ( 1 5 2 ) n ) F_{n} = \frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2})^{n}-(\frac{1-\sqrt{5}}{2})^{n}) . Using the identity 1 + x + x 2 + . . . + x n = x n + 1 1 x 1 1+x+x^{2}+...+x^{n} = \frac{x^{n+1}-1}{x-1} we can express k n k_n in terms of powers of 1 + 5 2 \frac{1+\sqrt{5}}{2} and 1 5 2 \frac{1-\sqrt{5}}{2} . Expansion will give you the first equation.)

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