Can someone help me with this:
pick $a_0 >0$ 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 $a_n>0$.
By AM-GM $$a_{n+1}=\frac{2}{3}a_n+\frac{1}{a_n}\geq2\sqrt{\frac{2}{3}a_n\cdot\frac{1}{a_n}}=\frac{2\sqrt2}{\sqrt3}>\frac{\sqrt{3}}{2}.$$ If $a_0>\sqrt3$ we obtain: $$a_{n+1}-a_n=\frac{(\sqrt3-a_n)(\sqrt3+a_n)}{3a_n}<0$$ because $$a_{n+1}-\sqrt3=\frac{(a_n-\sqrt3)(2a_n-\sqrt3)}{3a_n}>0$$ by induction.
Thus, $a$ is a decreasing and $a_n>\sqrt3$.
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 $0<a_0\leq\sqrt3$.
Hint: If a recursive sequence $a_{n+1}=f(a_n)$ converges, it must converge to a fixed point of $f$, that is, a number with $f(x)=x$.
(Of course, if you're being formal you need to prove that the sequence actually does converge before you can use the above.)
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 $a_{0}>0$.
You can prove that the sequences converges by proving that it is a Cauchy sequence, for example.
