RPS #011 - Celebrating The Best 5 National Competition

Calculus Level 3

If F n F_n defined as the n t h n^{th} term of Fibonacci sequence, find the value of lim n F n F n 1 \displaystyle \lim_{n \rightarrow \infty} \frac{F_n}{F_{n-1}}

Round your answer to 3 decimal places


The answer is 1.618.

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1 solution

Chew-Seong Cheong
Oct 26, 2016

Let the limit, if it exists, be λ = lim n F n F n 1 \displaystyle \lambda = \lim_{n \to \infty} \frac {F_n}{F_{n-1}} . As F n = F n 1 + F n 2 F_n = F_{n-1} + F_{n-2} , since F k + 1 > F k F_{k+1} > F_k for k 2 k \ge 2 , we have F n 1 < F n < 2 F n 1 F_{n-1} < F_n < 2F_{n-1} 1 < λ < 2 \implies 1 < \lambda < 2 . Since λ \lambda is bounded, by Bolzano-Weierstrass theorem, the limit exists.

λ = lim n F n F n 1 = lim n F n + 1 F n = lim n F n + F n 1 F n = 1 + 1 λ \begin{aligned} \lambda & = \lim_{n \to \infty} \frac {F_n}{F_{n-1}} \\ & = \lim_{n \to \infty} \frac {F_{n+1}}{F_n} \\ & = \lim_{n \to \infty} \frac {F_n + F_{n-1}}{F_n} \\ & = 1 + \frac 1\lambda \end{aligned}

λ 2 λ 1 = 0 \begin{aligned} \implies \lambda^2 - \lambda - 1& = 0 \end{aligned}

λ = 1 + 5 2 = φ 1.618 \begin{aligned} \implies \lambda = \dfrac {1 + \sqrt 5}2 = \varphi \approx \boxed{1.618} \end{aligned}

If the limit exists then this shows that it is the golden ratio, but for sake of completeness we should initially prove that a limit indeed exists. This can be done by first noting that F n 1 < F n < 2 F n 1 1 < λ < 2 F_{n-1} \lt F_{n} \lt 2F_{n-1} \Longrightarrow 1 \lt \lambda \lt 2 and then applying Bolzano-Weierstrass.

Brian Charlesworth - 4 years, 7 months ago

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Thanks, I learned some thing new today.

Chew-Seong Cheong - 4 years, 7 months ago

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