Golden Ratio Limit

Algebra Level 3

Find 2 a + 1 2a+1 :

lim x 0 ( n = 1 1008 ϕ n x 1008 ) 1 / x = F a + 1 ϕ + F a \large \lim_{x \to 0} {\left(\frac { \sum_{n=1}^{1008}\phi^{nx}}{1008}\right)}^{1/x} =\sqrt{F_{a+1}\cdot \phi +F_a}

where ϕ = 1 + 5 2 \phi=\frac{1+\sqrt{5}}2 and F m F_m is the m m th Fibonacci number with F 1 = 1 , F 2 = 1 F_1=1,F_2=1 and F m = F m 1 + F m 2 F_m=F_{m-1}+F_{m-2} .


The answer is 2017.

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1 solution

James Wilson
Dec 13, 2017

L = lim x 0 ( n = 1 1008 ϕ n x 1008 ) 1 / x ln L = lim x 0 ln n = 1 1008 ϕ n x ln 1008 x L=\lim_{x\rightarrow 0} \Big(\frac{\sum_{n=1}^{1008}\phi^{nx}}{1008}\Big)^{1/x}\Rightarrow \ln{L}=\lim_{x\rightarrow 0} \frac{\ln{\sum_{n=1}^{1008}\phi^{nx}}-\ln{1008}}{x} Since both the numerator and denominator approach zero, as x x approaches zero, I can apply L'Hospital's rule. ln L = lim x 0 n = 1 1008 n ϕ n x ln ϕ n = 1 1008 ϕ n x 1 \ln{L}=\lim_{x\rightarrow 0} \frac{\frac{\sum_{n=1}^{1008}n\phi^{nx}\ln{\phi}}{\sum_{n=1}^{1008}\phi^{nx}}}{1} Substituting x = 0 x=0 , ln L = n = 1 1008 n ln ϕ 1008 = ( 1008 ) ( 1009 ) 2 ln ϕ 1008 = 1009 2 ln ϕ \ln{L}=\frac{\sum_{n=1}^{1008}n\ln{\phi}}{1008}=\frac{\frac{(1008)(1009)}{2}\ln{\phi}}{1008}=\frac{1009}{2}\ln{\phi} L = exp ( 1009 2 ln ϕ ) = ϕ 1009 = F 1009 ϕ + F 1008 \Rightarrow L=\exp\Big(\frac{1009}{2}\ln{\phi}\Big)=\sqrt{\phi^{1009}}=\sqrt{F_{1009}\phi+F_{1008}}

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