Chapter 9 - Infinite Series - 9.2 Monotone Sequences - Exercises Set 9.2 - Page 614: 29

(a) ${x_1} = 60$, ${\ \ \ }$ ${x_2} = \dfrac{{1500}}{7} \approx 214.29$, ${\ \ \ }$ ${x_3} = \dfrac{{3750}}{{13}} \approx 288.46$, ${\ \ \ }$ ${x_4} = \dfrac{{75000}}{{251}} \approx 298.81$ (b) We show that $0 \lt {x_n} \lt 300$ for $n \ge 1$. (c) We show that $\left\{ {{x_n}} \right\}$ is increasing. (d) We show that $\left\{ {{x_n}} \right\}$ converges and its limit is $L=300$.

With $R=10$ and $K=300$, the Beverton-Holt model becomes ${x_{n + 1}} = \dfrac{{3000{x_n}}}{{300 + 9{x_n}}}$ (a) With ${x_1} = 60$ and using the formula ${x_{n + 1}} = \dfrac{{3000{x_n}}}{{300 + 9{x_n}}}$, we obtain ${x_1} = 60$, ${\ \ \ }$ ${x_2} = \dfrac{{1500}}{7} \approx 214.29$, ${\ \ \ }$ ${x_3} = \dfrac{{3750}}{{13}} \approx 288.46$, ${\ \ \ }$ ${x_4} = \dfrac{{75000}}{{251}} \approx 298.81$ (b) If $0 \lt {x_n} \lt 300$, then ${x_{n + 1}} = \dfrac{{3000{x_n}}}{{300 + 9{x_n}}} \lt \dfrac{{3000 \times 300}}{{300 + 9 \times 300}} = 300$ ${x_{n + 1}} \lt 300$ Thus, we conclude that $0 \lt {x_n} \lt 300$ for $n \ge 1$. This means that $300$ is an upper bound of $\left\{ {{x_n}} \right\}$. (c) Examine ${x_{n + 1}} - {x_n}$. ${x_{n + 1}} - {x_n} = \dfrac{{3000{x_n}}}{{300 + 9{x_n}}} - {x_n} = \dfrac{{3000{x_n} - 300{x_n} - 9{x_n}^2}}{{300 + 9{x_n}}} = \dfrac{{2700{x_n} - 9{x_n}^2}}{{300 + 9{x_n}}}$ ${x_{n + 1}} - {x_n} = \dfrac{{{x_n}\left( {2700 - 9{x_n}} \right)}}{{300 + 9{x_n}}}$ Since $\left\{ {{x_n}} \right\}$ are positive, the sign of ${x_{n + 1}} - {x_n}$ depends on the sign of $2700 - 9{x_n}$. For $0 \lt {x_n} \lt 300$, we have $2700 - 9{x_n} \gt 0$. Thus, ${x_{n + 1}} - {x_n} \gt 0$ ${x_{n + 1}} \gt {x_n}$ This implies that $\left\{ {{x_n}} \right\}$ is increasing. The figure attached confirm this result. (d) In part (b), we have shown that the sequence has an upper bound $300$. In part (c), we have shown that $\left\{ {{x_n}} \right\}$ is increasing. Thus, by 9.2.3 Theorem, the sequence converges. Next, we find its limit $L$. We have $L = \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}$ $L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{3000{x_n}}}{{300 + 9{x_n}}} = \dfrac{{3000\mathop {\lim }\limits_{n \to \infty } {x_n}}}{{300 + 9\mathop {\lim }\limits_{n \to \infty } {x_n}}} = \dfrac{{3000L}}{{300 + 9L}}$ $300L + 9{L^2} = 3000L$ $9{L^2} - 2700L = 0$ $9L\left( {L - 300} \right) = 0$ $L=0$, ${\ \ \ \ \ }$ $L=300$ Since ${x_1} = 60$ and the sequence is increasing, we must have $L=300$. Hence, $\mathop {\lim }\limits_{n \to \infty } {a_n} = 300$.