<s>[INST] 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!
 [/INST] 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 $L \ge \sqrt{3}$.
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