# Find the limit of $U_{n+1}=\frac13 (U_n+U_{n-1}^2+\frac59)$

Given that $$\begin{cases} U_1=U_2=0, \\ U_{n+1}=\frac13 (U_n+U_{n-1}^2+\frac59), \forall n\ge2 \end{cases}$$

Find the limit of $$U_n$$

My transformations:

The second condition yields: \begin{align} 3U_{n+1}=U_n+U_{n-1}^2+\frac59 \end{align}

So I assume that $$L$$ is the limit, then I solve the function for $$L$$, which yields $$\frac53$$ or $$\frac13$$

I also got $$U_3=\frac5{27} < \frac13$$ , $$U_4=\frac{20}{81} <\frac13$$, so I suspect that all numbers should be less than $$\frac13$$. Which then I did like:

If $$U_n$$ is the first one to exceed 1/3, then either $$U_{n-1}$$ or $$U_n$$ must exceed $$\frac13$$, which is inherently proven using induction as I have been mentioning earlier.

I just want to check if this is correct. And otherwise, how can it be solved differently?

Any review is appreciated!

• It is true by induction that $0\le U_n\le \frac{1}{3}$, then you could perhap use the transform $U_n = \frac{1}{3} - V_n$. Commented Nov 28, 2021 at 8:22
• Or you could prove by induction that $U_{n+1} - U_n \ge 0$. Commented Nov 28, 2021 at 8:29
• ok, with your argument $L=1/3$ or $L=5/3$ that is right, but first you have to prove that the limit exists, otherwise you can't use this argument Commented Nov 28, 2021 at 8:30
• @Marcos But it is an increasing sequence with an upper bound right? Commented Nov 28, 2021 at 8:35
• Yeah, that's it but you have to say it, just to make things rigurous. Commented Nov 28, 2021 at 8:38

Your equation $$\begin{cases} U_1=U_2=0, \\ U_{n+1}=\frac13 (U_n+U_{n-1}^2+\frac59), \forall n\ge2 \end{cases}$$ is equivalent to the map $$\begin{cases} x_{n+1}=y_n, \\ y_{n+1}=\frac13 (y_n+x_{n}^2+\frac59) \end{cases}$$ This map has fixed points $$p_1=(\frac{5}{3},\frac{5}{3})$$ and $$p_2=(\frac{1}{3},\frac{1}{3})$$. The jacobian matrix around $$p_2$$ has eigenvalues $$\lambda_1=\frac{2}{3}$$ and $$\lambda_2=-\frac{1}{3}$$. As $$|\lambda_1|<1$$ and $$|\lambda_2|<1$$, $$p_2$$ is an stable fixed point. You can show that all points with $$f(x,y)=(x-\frac{1}{3})^2+\frac{100}{64}(y-\frac{1}{3})^2<1$$ are in the basin of attraction of $$p_2$$. Hint: you can study what this function in a point before and after the action of the map.
First, we prove by induction that $$U_n\geq 0$$ for all $$n\geq 1$$. The base case is $$U_1=U_2=0$$, which is true by definition. Now, assume as induction hypothesis that $$U_n\geq 0$$, then$$U_{n+1}=\dfrac{1}{3}\left (U_n+U_{n-1}^2+\frac{5}{9}\right )\geq 0$$which finishes the induction.
Now, as you noted, we will prove that $$U_n\leq \dfrac{1}{3}$$ for all $$n\geq 1$$, using complete induction.
The base case is $$U_1=U_2=0\leq \dfrac{1}{3}$$, which is true by definition. Now, assume as induction hypothesis that $$U_n\leq \dfrac{1}{3}$$ for all $$n\leq k$$, then for $$k+1$$ we have$$U_{k+1}=\frac{1}{3}\left (U_k+U_{k-1}^2+\frac{5}{9}\right )\leq \frac{1}{3}\left (\frac{1}{3}+\frac{1}{9}+\frac{5}{9}\right )=\frac{1}{3}$$which finishes the induction.
Then we have that $$0\leq U_n\leq \dfrac{1}{3}$$ for all $$n\geq 1$$.
Next, we prove by induction that $$U_n\leq U_{n+1}$$ for all $$n\geq 1$$. The base case is $$U_1=0\leq 0=U_2$$ which is true by definition. Now, assume as inducing hypothesis that $$U_{n-1}\leq U_n$$ for $$n\geq 2$$, first observe that from $$0\leq U_{n-1}\leq U_n\leq \dfrac{1}{3}$$ it follows that $$U_{n-1}^2\leq U_n^2\leq U_n$$, hence$$2U_n-U_{n-1}^2-\frac{5}{9}\leq 2U_n-U_n-\frac{5}{9}=U_n-\frac{5}{9}\leq 0$$from this follows that$$3U_n\leq U_n+U_{n-1}^2+\frac{5}{9}$$which implies that$$U_n\leq \frac{1}{3}\left (U_n+U_{n-1}+\frac{5}{9}\right )=U_{n+1}$$as wanted. This finished the induction.
Therefore, $$\{U_n\}$$ is an increasing bounded sequence, hence it is a convergent sequence. Let $$L=\lim \limits _{n\to \infty}U_n$$. Then\begin{align*}L & =\lim \limits _{n\to \infty}U_n \\ & =\lim \limits _{n\to \infty}U_{n+1} \\ & =\lim \limits _{n\to \infty}\frac{1}{3}\left (U_n+U_{n-1}^2+\frac{5}{9}\right ) \\ & =\frac{1}{3}\left (L+L^2+\frac{5}{9}\right ). \end{align*}From this follows that $$L=\dfrac{1}{3}$$ or $$L=\dfrac{5}{3}$$. But $$L\leq 1$$ because $$U_n\leq 1$$ for all $$n\geq 1$$, hence $$L=\dfrac{1}{3}$$, i.e. $$\lim \limits _{n\to \infty}U_n=\dfrac{1}{3}$$.