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\noindent {\bf Question}

\noindent The {\bf Fibonacci sequence} $\{ a_n\: |\: n\ge 0\}$ is
formed by setting $a_0 =0$, $a_1 =1$, and $a_n =a_{n-1} + a_{n-2}$
for $n\ge 2$.  Consider the derived sequence $\{ q_n
=\frac{a_n}{a_{n-1}}\}$ of quotients of consecutive terms of the
Fibonacci sequence.  Show that if $\lim_{n\rightarrow\infty} q_n$
exists, then $\lim_{n\rightarrow\infty} q_n
=\frac{1+\sqrt{5}}{2}$.
\medskip

\noindent {\bf Answer}

\noindent Suppose that $\{ q_n\}$ converges and set $x
=\lim_{n\rightarrow\infty} q_n$.  Now, note that
\[ q_n =\frac{a_n}{a_{n-1}} =\frac{a_{n-1} +a_{n-2}}{a_{n-1}}
=1+\frac{a_{n-2}}{a_{n-1}} =1 + \frac{1}{q_{n-1}}. \] Hence,
\[ x =\lim_{n\rightarrow\infty} q_n =\lim_{n\rightarrow\infty} \left(
1+\frac{1}{q_{n-1}} \right) =1+\frac{1}{\lim_{n\rightarrow\infty}
q_{n-1}} =1+\frac{1}{x}, \] since $\lim_{n\rightarrow\infty}
q_{n-1} =x$ as well.  Therefore, $x =1+\frac{1}{x}$, and so
(multiplying through by $x$ and simplifying) $x$ satisfies the
quadratic equation $x^2 -x-1=0$.  By the quadratic formula, this
yields that $x =\frac{1}{2}\left(1 \pm\sqrt{5} \right)$.  However,
since $q_n \ge 0$ for all $n$, it must be that $x\ge 0$ as well,
and so $x =\frac{1}{2}\left(1 +\sqrt{5}\right)$.

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