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Find a limit of sequence: $$a_{n+2}=a^2_{n+1}+\frac{1}{6}\cdot a_n+\frac{1}{9}$$ $$a_1=0,a_2=0$$

I tried to prove that $a_n$ is bounded and monotonic, but I couldn't prove that $a_n$ is monotonic (by strong induction). So please give a idea how to show that $a_n$ is monotonic.

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(this proof takes a little longer, but gives a neat bound on the rate of convergence)

  1. If you want a universal bound of the form $a_n\le M$, the recursion tells you that you're going to need $M\le M^2+\frac{M}{6}+\frac{1}{9}$, which works out to be equivalent to $(6M-1)(3M-2)\le0$. The lowest bound this allows us to take is $M=\frac{1}{6}$, i.e. that $a_n\le\frac{1}{6}$ for all $n$. Feeding this back into the recursion, we see that this inequality does indeed hold.

  2. If you're hoping for a limit $L$ to exist, the recursion tells you that you're going to need $L=L^2+\frac{L}{6}+\frac{1}{9}$, and similar reasoning to before gives that the only possible limits are $\frac{1}{6}$ and $\frac{2}{3}$. But having established our earlier bounds on $a_n$, we see that if a limit does exist, it will necessarily be $\frac{1}{6}$.

  3. With the aim of showing that $a_n\to\frac{1}{6}$, we set $a_n=\frac{1}{6}-b_n$ and work through the algebra to obtain the recurrence $b_{n+2}=\frac{b_n}{6}+\frac{b_{n+1}}{3}-b_{n+1}^2$. Our earlier work gives that $0<b_n\le \frac{1}{6}$ for all $n$, and some examination hints that this $b_n$ decays to 0 geometrically, so we seek a bound of the form $b_n\le Cr^n$. Appealing to a strong-induction-style argument, we would need:

$$b_{n+2}=\frac{b_n}{6}+\frac{b_{n+1}}{3}-b_{n+1}^2\le\frac{Cr^n}{6}+\frac{Cr^{n+1}}{3}=Cr^{n+2}\left(\frac{1}{6r^2}+\frac{1}{3r}\right)$$

So, if we can make $\left(\frac{1}{6r^2}+\frac{1}{3r}\right)\le 1$, we can get this inequality to hold (for a suitable $C$) by strong induction. $r=\frac{1+\sqrt7}{6}$ is the smallest $r$ which works for this, and we note that $r\le1$, guaranteeing convergence.

Thus, we see that $0<b_n<\frac{r^{n-2}}{6}$, and hence $b_n\to 0, a_n\to \frac{1}{6}$.

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$ a_3-a_2\ge 0 $. Assume, $ a_{k+1}-a_k\ge0 $ for every $ k\le n $ then, $ a_{n+2}-a_{n+1}= a_{n+1}^2-a_n^2+(a_{n}-a_{n-1})/6 $. which is greater or equal to 0 since $ a^2-b^2=(a+b)(a-b) $ and by the assumption. This settles the monotonic part.

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  • $\begingroup$ Then prove by induction that all terms are $\le 2/3$. $\endgroup$ – GEdgar Feb 6 '16 at 13:12
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$$a_{n+2}≥\frac19$$ $x^2-\frac56x+\frac19=0$ This equation's roots are $\frac16, \frac23$. By $a_1=a_2=0$ converge to$\frac16$. $$\frac16>a_{n}≥\frac19$$

and when $a_{k+1}≥a_{k},a_{k+2}≥a_{k+1},$

$$a_{k+3}-a_{k+2}=a^2_{k+2}+\frac{1}{6}\cdot a_{k+1}+\frac{1}{9}-a_{k}$$ $$≥a^2_{k}+\frac16a_{k}+\frac19-a_k ≥(a_k-\frac5{12})^2-\frac1{16}≥0$$ ∴$a_n≥a_{n-1}$

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