Sequence converging to $\sqrt{3}$ Can someone help me with this:

pick $a_0 &gt;0$ and define {$a_n$} by: $a_{n+1} = 2a_n/3 + 1/a_n.$

Explain why the sequence converges to $\sqrt{(3)}.$

I have calculated the first 10 elements and I can see that it's getting closer to $\sqrt{(3)}\approx1.7320$
It's obvious that $a_n&gt;0$.

By AM-GM $$a_{n+1}=\frac{2}{3}a_n+\frac{1}{a_n}\geq2\sqrt{\frac{2}{3}a_n\cdot\frac{1}{a_n}}=\frac{2\sqrt2}{\sqrt3}&gt;\frac{\sqrt{3}}{2}.$$
If $a_0&gt;\sqrt3$ we obtain:
$$a_{n+1}-a_n=\frac{(\sqrt3-a_n)(\sqrt3+a_n)}{3a_n}&lt;0$$
because $$a_{n+1}-\sqrt3=\frac{(a_n-\sqrt3)(2a_n-\sqrt3)}{3a_n}&gt;0$$
by induction.

Thus, $a$ is a decreasing and $a_n&gt;\sqrt3$.

Id est, there is a $\lim\limits_{n\rightarrow+\infty}a_n$.

Let $\lim\limits_{n\rightarrow+\infty}a_n=a$.

Thus, $a=\frac{2}{3}a+\frac{1}{a}$, which gives $a=\sqrt3$.

By the same way we can get that $a=\sqrt3$ for $0&lt;a_0\leq\sqrt3$.
You first have to prove that it converges, if it converges it must converge to a fixed point:
$L=\frac{2}{3}L+\frac{1}{L}$

solving it you have, multiplying by $3L$

$3L^{2}=2L^{2}+3$

that leads to

$L^{2}=3\Rightarrow L=\sqrt{3}$ $ \quad$because your sequence is a sequence of positive numbers, because of $a_{0}&gt;0$.

You can prove that the sequences converges by proving that it is a Cauchy sequence, for example.