Problem Statement

Let \(F_n\) be the Fibonacci sequence. Compute:

\[\lim_{n \to \infty} \frac{F_{n+2}}{F_n}.\]

Your answer should be in the form:

\[\frac{a + \sqrt{b}}{c},\]

where \(a, b, c\) are pairwise relatively prime. Find \(abc\).

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Solution

We begin with the Fibonacci recurrence:

\[F_{n+2} = F_{n+1} + F_n\]

So the expression becomes:

\[\frac{F_{n+2}}{F_n} = \frac{F_{n+1} + F_n}{F_n} = \frac{F_{n+1}}{F_n} + 1\]

Let:

\[x_n = \frac{F_{n+1}}{F_n}\]

Then:

\[\frac{F_{n+2}}{F_n} = x_n + 1\]

We now analyze the behavior of \(x_n\) as \(n \to \infty\). Using the Fibonacci recurrence again:

\[x_n = \frac{F_{n+1}}{F_n} = \frac{F_n + F_{n-1}}{F_n} = 1 + \frac{F_{n-1}}{F_n}\]

As \(n \to \infty\), one can obviously safely assume that \(\frac{F_{n-1}}{F_n} \to \frac{1}{x_n}\)

So we get:

\[x = 1 + \frac{1}{x}\]

Multiply both sides by \(x\):

\[x^2 = x + 1 \quad \Rightarrow \quad x^2 - x - 1 = 0\]

Solving this quadratic:

\[x = \frac{1 \pm \sqrt{5}}{2}\]

We discard the negative root (since Fibonacci ratios are positive), so:

\[x = \frac{1 + \sqrt{5}}{2}\]

Therefore,

\[\lim_{n \to \infty} \frac{F_{n+2}}{F_n} = x + 1 = \frac{1 + \sqrt{5}}{2} + 1 = \frac{3 + \sqrt{5}}{2}\]

So the required expression is:

\[\frac{a + \sqrt{b}}{c} = \frac{3 + \sqrt{5}}{2}\]

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Final Answer: \(a = 3, b = 5, c = 2 \Rightarrow abc = 3 \times 5 \times 2 = \boxed{30}\)

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