Chapter 9 - Infinite Series - 9.1 Sequences - Exercises Set 9.1 - Page 607: 47

(a) We show that $\dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = 1 + \dfrac{{{a_n}}}{{{a_{n + 1}}}}$, for $n \ge 1$ (b) We give an informal argument that $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = L$ (c) We show that $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{1}{2}\left( {1 + \sqrt 5 } \right)$

(a) We have the Fibonacci sequence: $1,1,2,3,5,8,13,21,...$ The general term is given by ${a_{n + 2}} = {a_{n + 1}} + {a_n}$, ${\ \ \ \ \ }$ for $n \ge 1$ Dividing both sides by ${a_{n + 1}}$ gives $\dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = 1 + \dfrac{{{a_n}}}{{{a_{n + 1}}}}$, for $n \ge 1$ (b) Consider the two limits: $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}}$ and $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}}$. For $n \to + \infty $, both $n + 1 \to \infty $ and $n + 2 \to \infty $ are true. Therefore, if $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = L$, then $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = L$ So, $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = L$ (c) From part (b), we get: $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = L$ From part (a), we know that $\dfrac{{{a_{n + 2}}}}{{{a_{n + 1}}}} = 1 + \dfrac{{{a_n}}}{{{a_{n + 1}}}}$. Thus, $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = \mathop {\lim }\limits_{n \to + \infty } \left( {1 + \dfrac{{{a_n}}}{{{a_{n + 1}}}}} \right) = L$ That is, $1 + \mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_n}}}{{{a_{n + 1}}}} = L$ Since $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_n}}}{{{a_{n + 1}}}} = \mathop {\lim }\limits_{n \to + \infty } \dfrac{1}{{\dfrac{{{a_{n + 1}}}}{{{a_n}}}}} = \dfrac{1}{{\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}}}}$, so $1 + \dfrac{1}{{\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}}}} = L$ We have $\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_{n + 1}}}}{{{a_n}}} = L$. Thus, $1 + \dfrac{1}{L} = L$ ${L^2} - L - 1 = 0$ $L = \dfrac{{1 \pm \sqrt {{{\left( { - 1} \right)}^2} + 4} }}{2} = \dfrac{1}{2}\left( {1 \pm \sqrt 5 } \right)$ $L = \dfrac{1}{2}\left( {1 - \sqrt 5 } \right)$, ${\ \ \ \ \ }$ $L = \dfrac{1}{2}\left( {1 + \sqrt 5 } \right)$ Since the Fibonacci sequence is positive, we take $L = \dfrac{1}{2}\left( {1 + \sqrt 5 } \right)$.