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!

  • $\begingroup$ It is true by induction that $0\le U_n\le \frac{1}{3}$, then you could perhap use the transform $U_n = \frac{1}{3} - V_n$. $\endgroup$ Nov 28, 2021 at 8:22
  • $\begingroup$ Or you could prove by induction that $U_{n+1} - U_n \ge 0$. $\endgroup$ Nov 28, 2021 at 8:29
  • $\begingroup$ 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 $\endgroup$
    – Marcos
    Nov 28, 2021 at 8:30
  • $\begingroup$ @Marcos But it is an increasing sequence with an upper bound right? $\endgroup$ Nov 28, 2021 at 8:35
  • $\begingroup$ Yeah, that's it but you have to say it, just to make things rigurous. $\endgroup$
    – Marcos
    Nov 28, 2021 at 8:38

2 Answers 2


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.


I will give an elementary solution.

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. $\lim \limits _{n\to \infty}U_n=\dfrac{1}{3}$.


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