Convergence of sequence dependent on initial values Hi I have searched alot of questions but I can't find anything that is close to what I am trying to do.
I have a sequence $a_{n}= \frac{2a_{n-1}}{3}+\frac{64}{3a_{n-1}}, n=1,2,3...$ 
I am trying to locate all initial values $a_1\in\mathbb{R}$ such that the sequence converges. I'm kinda lost so any help would be appreciated. 
 A: Suppose that the sequence converges.  Then
$$a = {2a\over 3}+{64\over 3a}\\
3a^2 = 2a^2+64\\
a^2 = 64$$
This yields two values that the sequence converges to.  Next we need to find out if there is a time when $a_{n+1}$ is moving away from either convergence point instead of towards it.  From the initial equation, $a_0 = 0$ is not allowed, so we have four areas to search in: $(-\infty,-8)\cup(-8,0)\cup(0,8)\cup(8,\infty)$.  We can even map $(-q,0)\cup(0,r)$ to $(-\infty,-8)\cup(8,\infty)$ for some values $q,r\in(0, 8)$ by setting $a_0 = q,r$, and narrow our two inner search regions.  Because the equation is symmetric about $0$, we can see that $q=r$.  Trying a few values, we find that $a_0=4$ produces $a_1 = 8$, and therefore $q = 4$.
Now we try two values: $a_0 = 8+k,8-m$ for $k\gt 0, m\in(0,4)$:
$$a_1 = \frac 23(8+k)+\frac{64}{24+3k}\\
={2(8+k)^2+64\over 3(8+k)}\lt{3(8+k)^2\over 3(8+k)}=8+k\\
{2(8+k)^2+64\over 3(8+k)}={192+32k+k^2\over 3(8+k)}\gt {24(8+k)\over 3(8+k)}=8$$
Therefore $a_0\in(-\infty,-8)\cup(-4,0)\cup(0,4)\cup(8,\infty)$ will produce convergence.
Now we consider $a_0 = 8-m$:
$$a_1={2(8-m)\over 3}+{64\over 3(8-m)}={2(8-m)^2+64\over 3(8-m)}\\
{2(8-m)^2+64\over 3(8-m)}\gt {3(8-m)^2\over 3(8-m)}=8-m\\
{2(8-m)^2+64\over 3(8-m)}={192-32m+m^2\over 3(8-m)}\lt {24(8-m)\over 3(8-m)}=8$$
Therefore, $a_0\in(-8,-4)\cup(4,8)$ will produce convergence.  Putting all of the initial value sets together, we get that $a_0\in(-\infty,0)\cup(0,\infty)$ will produce convergence, negative values to $-8$, positive values to $8$.
