Chapter 11 - Infinite Series - Chapter Review Exercises - Page 591: 26

(a) $\left( {{a_2},{a_3},{a_4},{a_5}} \right) \approx \left( {2.82843,2.97127,2.99521,2.9992} \right)$ (b) We show that $\left\{ {{a_n}} \right\}$ is increasing and bounded by $3$. (c) We prove $\mathop {\lim }\limits_{n \to \infty } {a_n}$ exists and its value: $\mathop {\lim }\limits_{n \to \infty } {a_n} = 3$

(a) We have ${a_{n + 1}} = \sqrt {{a_n} + 6} $ with ${a_1} = 2$. So, ${a_2} = \sqrt {2 + 6} = 2\sqrt 2 $ ${a_3} = \sqrt {2\sqrt 2 + 6} $ ${a_4} = \sqrt {\sqrt {2\sqrt 2 + 6} + 6} $ ${a_5} = \sqrt {\sqrt {\sqrt {2\sqrt 2 + 6} + 6} + 6} $ So, $\left( {{a_2},{a_3},{a_4},{a_5}} \right) \approx \left( {2.82843,2.97127,2.99521,2.9992} \right)$ (b) We have the sequence: ${a_1} = 2$, ${\ \ \ }$ ${a_2} = \sqrt {{a_1} + 6} $, ${\ \ \ }$ ${a_3} = \sqrt {{a_2} + 6} $, ${\ \ \ }$ ${a_4} = \sqrt {{a_3} + 6} $, ${\ \ }$ $\cdot\cdot\cdot$ Since ${a_1} > 0$, so ${a_1} < {a_2}$, ${a_2} < {a_3}$, $\cdot\cdot\cdot$ Therefore, ${a_1} < {a_2} < {a_3} < {a_4} < \cdot\cdot\cdot$ We have ${a_1} = 2$ and ${a_{n + 1}} = \sqrt {{a_n} + 6} $. Since ${a_1} < 3$, so ${a_2} < \sqrt {3 + 6} = 3$. Since ${a_2} < 3$, so ${a_3} < \sqrt {3 + 6} = 3$. Since ${a_3} < 3$, so ${a_4} < \sqrt {3 + 6} = 3$. And so on, we have all terms ${a_n} < 3$. This implies that ${a_n}$ is bounded by $3$. Hence, $\left\{ {{a_n}} \right\}$ is increasing and bounded by $3$. (c) We know from part (b) that $\left\{ {{a_n}} \right\}$ is increasing and bounded by $3$. Thus, by Theorem 6 of Section 11.1: $\left\{ {{a_n}} \right\}$ converges, that is, $\mathop {\lim }\limits_{n \to \infty } {a_n}$ exists. In this case, let $M=3$. Now, we evaluate $\mathop {\lim }\limits_{n \to \infty } {a_n}$. Let $L = \mathop {\lim }\limits_{n \to \infty } {a_n}$. By definition, we also have $L = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}$ Since ${a_{n + 1}} = \sqrt {{a_n} + 6} $, so $L = \mathop {\lim }\limits_{n \to \infty } \sqrt {{a_n} + 6} $ $L = \sqrt {\mathop {\lim }\limits_{n \to \infty } {a_n} + 6} $ But $L = \mathop {\lim }\limits_{n \to \infty } {a_n}$, so $L = \sqrt {L + 6} $ ${L^2} = L + 6$ ${L^2} - L - 6 = 0$ $\left( {L + 2} \right)\left( {L - 3} \right) = 0$ $L=-2$, ${\ \ \ \ \ }$ $L=3$ Since ${a_n} > 0$, we choose $L=3$. Thus, $L = \mathop {\lim }\limits_{n \to \infty } {a_n} = 3$.