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!

$$\frac{a_{n+1}+\sqrt3}{a_{n+1}-\sqrt3} = \frac{a_n+\frac{1}{a_n}+2\sqrt3}{a_n+\frac{1}{a_n}-2\sqrt3}=\left(\frac{a_n+\sqrt3}{a_n-\sqrt3}\right)^2 \\\implies \frac{a_n+\sqrt3}{a_n-\sqrt3}=\left(\frac{a_1+\sqrt3}{a_1-\sqrt3}\right)^{2^{n-1}}= \left(\frac{1+\sqrt3}{1-\sqrt3}\right)^{2^{n-1}}\\ \implies 
a_n = \sqrt 3 \cdot \frac{(1+\sqrt3)^{2^{n-1}}+ (1-\sqrt3)^{2^{n-1}}}{(1+\sqrt3)^{2^{n-1}}-(1-\sqrt3)^{2^{n-1}}} \to \sqrt 3, n \to \infty.$$
