Prove the following sequence is convergent: $a(1) = 1$ and $a(n+1) = 1/2 \cdot (a(n) + 3/a(n))$ I know the limit is $\sqrt 3$. And I tried proving that the sequence is decreasing from $a(2)$, yet when I got to this point, I have been stuck:
$a(n+1) - a(n) = (3 - a(n)^2) / 2a(n)$ ?
Thank you very much!
 A: Let’ s prove by induction that for $n \ge 2$ $a_n -\sqrt 3 \gt0$
It’s true for $n=2$ as $a_2=2$. Suppose that the inequality holds for $n$. As
$$\begin{aligned}
a_{n+1} -\sqrt 3 &= \frac{1}{2}\left(a_n +\frac{3}{a_n}\right)-\frac{1}{2}\left(\sqrt 3+\frac{3}{\sqrt 3}\right)\\
&=\frac{1}{2}\left(a_n-\sqrt 3\right)\left(1+\frac{\sqrt 3}{a_n}\right)
\end{aligned}$$
The inequality also holds for $n+1$.
Now using the equality written in your question and the inequality we just proved, you get the desired result that $\{a_n\}$ is decreasing starting at the second term.
A: Hint:
On the interval $(\sqrt 3,+\infty)$ the function $\:f(x)=\dfrac12\Bigl(x+\dfrac 3x\Bigr)$ is increasing and $\:f(x)<x$.
Note that $a_2=2\in(\sqrt 3,+\infty)$.
A: $$\frac{a_{n+1}+\sqrt3}{a_{n+1}-\sqrt3} = \frac{a_n+\frac{1}{a_n}+2\sqrt3}{a_n+\frac{1}{a_n}-2\sqrt3}=\left(\frac{a_n+\sqrt3}{a_n-\sqrt3}\right)^2 \\\implies \frac{a_n+\sqrt3}{a_n-\sqrt3}=\left(\frac{a_1+\sqrt3}{a_1-\sqrt3}\right)^{2^{n-1}}= \left(\frac{1+\sqrt3}{1-\sqrt3}\right)^{2^{n-1}}\\ \implies 
a_n = \sqrt 3 \cdot \frac{(1+\sqrt3)^{2^{n-1}}+ (1-\sqrt3)^{2^{n-1}}}{(1+\sqrt3)^{2^{n-1}}-(1-\sqrt3)^{2^{n-1}}} \to \sqrt 3, n \to \infty.$$
A: Using algebra
Rewriting
$$a_{n+1}=\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)$$ as
$$a_{n+1}=a_n-\left(a_n-\frac{1}{2} \left(a_n+\frac{3}{a_n}\right)\right)=a_n-\frac{a_n^2-3}{2 a_n}$$ we can recognize the itegrative scheme for solving $a^2=3$; so the limit is $\sqrt 3$.
Considering the function $f(a)=a^2-3$, starting with $a_0=1$, we have
$f(1)=-2$ and $f''(1)=2$, $f(1)\times f''(1)=-4<0$, then, by Darboux theorem, we shall have one (and only one overshoot of the solution). This corresponds to $a_2$ and, later, we shall converge quadratically to the root.
$$\left(
\begin{array}{cc}
n & a_n \\
 1 & 1.0000000000000000000 \\
 2 & 2.0000000000000000000 \\
 3 & 1.7500000000000000000 \\
 4 & 1.7321428571428571429 \\
 5 & 1.7320508100147275405 \\
 6 & 1.7320508075688772953 \\
 7 & 1.7320508075688772935
\end{array}
\right)$$
A: A more direct proof.
$s_{n+1}
=\dfrac12(s_n+\dfrac{a}{s_n})
$.
$\begin{array}\\
s_{n+1}^2
&=\dfrac14(a_n^2+2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n^2-2a+\dfrac{a^2}{s_n^2})\\
&=a+\dfrac14(s_n-\dfrac{a}{s_n})^2\\
&\gt a\\
\end{array}
$
so
$s_n > \sqrt{a}
$.
$\begin{array}\\
s_{n+1}-\sqrt{a}
&=\dfrac12(s_n-2\sqrt{a}+\dfrac{a}{s_n})\\
&=\dfrac12(\sqrt{s_n}-\dfrac{\sqrt{a}}{\sqrt{s_n}})^2\\
&=\dfrac1{2s_n}(s_n-\sqrt{a})^2\\
\dfrac{s_{n+1}-\sqrt{a}}{s_n-\sqrt{a}}
&=\dfrac1{2s_n}(s_n-\sqrt{a})\\
&\lt \dfrac12\\
\end{array}
$
so $s_n$ converges to
$\sqrt{a}$.
A: Observe that $a_n > 0, \forall n \ge 1$, and by using the AM-GM inequality: $\forall n \ge 1 \implies (a_{n+1})^2 \ge \left(\dfrac{1}{2}\cdot 2\sqrt{a_n\cdot \dfrac{3}{a_n}}\right)^2 = \left(\sqrt{3}\right)^2 = 3 \implies a_{n+1} \ge \sqrt{3}, \forall n \ge 1$. Thus: $a_{n+1} - a_n = \dfrac{1}{2}\left(a_n+\dfrac{3}{a_n}\right)-a_n= \dfrac{1}{2}\left(\dfrac{3}{a_n}-a_n\right)= \dfrac{3-a_n^2}{2a_n}< 0\implies a_{n+1} < a_n$. The sequence is decreasing and bounded below by $\sqrt{3}$ hence converges to $L$ that satisfies the equation: $L = \dfrac{1}{2}\left(L+\dfrac{3}{L}\right)\implies L^2 = 3\implies L = \sqrt{3}$ since $L \ge \sqrt{3}$.
