Answer
(a) We show that ${a_n} \ge \sqrt 3 $ for $n \ge 2$.
(b) We show that $\left\{ {{a_n}} \right\}$ is eventually decreasing.
(c) We show that $\left\{ {{a_n}} \right\}$ converges and its limit is $L = \sqrt 3 $
Work Step by Step
(a) Let $f\left( n \right) = \dfrac{1}{2}\left( {{a_n} + \dfrac{3}{{{a_n}}}} \right)$. Write $x = {a_n}$. So,
$f\left( x \right) = \dfrac{1}{2}\left( {x + \dfrac{3}{x}} \right)$
Taking the derivative with respect to $x$ gives
$f'\left( x \right) = \dfrac{1}{2}\left( {1 - \dfrac{3}{{{x^2}}}} \right)$
Another derivative with respect to $x$ gives
$f{\rm{''}}\left( x \right) = \dfrac{3}{{{x^3}}}$
For $x \gt 0$, we have $f{\rm{''}}\left( x \right) \gt 0$. Thus, the function $f$ has a minimum. Solving the following equation
$f'\left( x \right) = \dfrac{1}{2}\left( {1 - \dfrac{3}{{{x^2}}}} \right) = 0$,
we obtain the minimum point $x = \sqrt 3 \approx 1.732$.
Since $n$ is integer, the closest integer to $1.732$ is $2$.
Hence, ${a_n} \ge \sqrt 3 $ for $n \ge 2$.
(b) Examine ${a_{n + 1}} - {a_n}$.
${a_{n + 1}} - {a_n} = \dfrac{1}{2}\left( {{a_n} + \dfrac{3}{{{a_n}}}} \right) - {a_n} = \dfrac{1}{2}\left( {\dfrac{3}{{{a_n}}} - {a_n}} \right)$
${a_{n + 1}} - {a_n} = \dfrac{1}{2}\left( {\dfrac{{3 - {a_n}^2}}{{{a_n}}}} \right)$
In part (a), we have shown that ${a_n} \ge \sqrt 3 $ for $n \ge 2$. Thus, $3 - {a_n}^2 \le 0$.
Since all terms are positive, it follows that ${a_{n + 1}} - {a_n} \le 0$. Namely,
${a_{n + 1}} \le {a_n}$, ${\ \ \ \ \ }$ for $n \ge 2$
Since ${a_n} \ge \sqrt 3 $ for $n \ge 2$ and at the same time ${a_{n + 1}} \le {a_n}$ for $n \ge 2$, we conclude that $\left\{ {{a_n}} \right\}$ is eventually decreasing. This fact is shown in the figure attached.
(c) In part (a), we have shown that the sequence has a lower bound $\sqrt 3 $. In part (b), we have shown that the sequence is eventually decreasing. Thus, by 9.2.4 Theorem, the sequence converges. Next, we find its limit $L$.
We have
$L = \mathop {\lim }\limits_{n \to \infty } {a_n} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}}$
$L = \mathop {\lim }\limits_{n \to \infty } \dfrac{1}{2}\left( {{a_n} + \dfrac{3}{{{a_n}}}} \right) = \dfrac{1}{2}\left( {\mathop {\lim }\limits_{n \to \infty } {a_n} + \dfrac{3}{{\mathop {\lim }\limits_{n \to \infty } {a_n}}}} \right)$
$L = \dfrac{1}{2}\left( {L + \dfrac{3}{L}} \right)$
${L^2} - 3 = 0$
$\left( {L - \sqrt 3 } \right)\left( {L + \sqrt 3 } \right) = 0$
$L = \sqrt 3 $, ${\ \ \ \ \ }$ $L = - \sqrt 3 $
Since all terms are positive, we must have $L = \sqrt 3 $. Hence,
$\mathop {\lim }\limits_{n \to \infty } {a_n} = \sqrt 3 $
