Answer
(a) We show that ${x_n} \gt K$ for all $n \ge 1$.
(b) We show that $\left\{ {{x_n}} \right\}$ is decreasing.
(c) We show that $\left\{ {{x_n}} \right\}$ converges and its limit is $L=K$.
Work Step by Step
(a) Recall the Beverton-Holt model:
${x_{n + 1}} = \dfrac{{RK{x_n}}}{{K + \left( {R - 1} \right){x_n}}}$
for some positive constants $R$ and $K$ with $R \gt 1$.
For ${x_1} \gt K$, we have
${x_2} = \dfrac{{RK{x_1}}}{{K + \left( {R - 1} \right){x_1}}} \gt \dfrac{{RK \times K}}{{K + \left( {R - 1} \right)K}} = \dfrac{{R{K^2}}}{{RK}} = K$
So, ${x_2} \gt K$.
For ${x_2} \gt K$, we have
${x_3} = \dfrac{{RK{x_2}}}{{K + \left( {R - 1} \right){x_2}}} \gt \dfrac{{RK \times K}}{{K + \left( {R - 1} \right)K}} = \dfrac{{R{K^2}}}{{RK}} = K$
So, ${x_3} \gt K$.
And so on, we obtain ${x_n} \gt K$ for all $n \ge 1$. This means that $K$ is a lower bound of $\left\{ {{x_n}} \right\}$.
Since $K \gt 0$, this also implies that $\left\{ {{x_n}} \right\}$ are positive.
(b) Examine ${x_{n + 1}} - {x_n}$.
${x_{n + 1}} - {x_n} = \dfrac{{RK{x_n}}}{{K + \left( {R - 1} \right){x_n}}} - {x_n} = \dfrac{{RK{x_n} - K{x_n} - \left( {R - 1} \right){x_n}^2}}{{K + \left( {R - 1} \right){x_n}}} = \dfrac{{K{x_n}\left( {R - 1} \right) - \left( {R - 1} \right){x_n}^2}}{{K + \left( {R - 1} \right){x_n}}}$
${x_{n + 1}} - {x_n} = \dfrac{{{x_n}\left[ {K\left( {R - 1} \right) - \left( {R - 1} \right){x_n}} \right]}}{{K + \left( {R - 1} \right){x_n}}}$
Since $\left\{ {{x_n}} \right\}$ are positive, the sign of ${x_{n + 1}} - {x_n}$ depends on the sign of $K\left( {R - 1} \right) - \left( {R - 1} \right){x_n}$.
Write
$K\left( {R - 1} \right) - \left( {R - 1} \right){x_n} = \left( {R - 1} \right)\left( {K - {x_n}} \right)$
With $R \gt 1$ and the result in part (a) that ${x_n} \gt K$ for all $n \ge 1$, we conclude that
$K\left( {R - 1} \right) - \left( {R - 1} \right){x_n} = \left( {R - 1} \right)\left( {K - {x_n}} \right) \lt 0$
Therefore, ${x_{n + 1}} - {x_n} \lt 0$, or
${x_{n + 1}} \lt {x_n}$
Hence, $\left\{ {{x_n}} \right\}$ is decreasing.
(c) In part (a), we have shown that the sequence has a lower bound $K$. In part (b), we have shown that $\left\{ {{x_n}} \right\}$ is decreasing. Thus, by 9.2.4 Theorem, the sequence converges. Next, we find its limit $L$.
We have
$L = \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}$
$L = \mathop {\lim }\limits_{n \to \infty } \dfrac{{RK{x_n}}}{{K + \left( {R - 1} \right){x_n}}} = \dfrac{{RK\mathop {\lim }\limits_{n \to \infty } {x_n}}}{{K + \left( {R - 1} \right)\mathop {\lim }\limits_{n \to \infty } {x_n}}} = \dfrac{{RKL}}{{K + \left( {R - 1} \right)L}}$
$KL + \left( {R - 1} \right){L^2} = RKL$
$\left( {R - 1} \right){L^2} - KL\left( {R - 1} \right) = 0$
$L\left( {R - 1} \right)\left( {L - K} \right) = 0$
$L\left( {R - 1} \right) = 0$, ${\ \ \ \ \ }$ $\left( {L - K} \right) = 0$
Since $R \gt 1$, we obtain the solutions:
$L=0$, ${\ \ \ \ \ }$ $L=K$
Since ${x_n} \gt K \gt 0$ for all $n \ge 1$, we must have $L=K$. Hence, $\mathop {\lim }\limits_{n \to \infty } {a_n} = K$.
