Finding $\limsup_{n\to\infty}\sqrt[n]{|a_n|}$ for a recursively defined sequence $(a_n)$ with $a_{n}=\frac13(a_{n-1}+a_{n-2}+a_{n-1}a_{n-2})$

Solution 1:

Hint:

$$\frac{a_n}{a_{n-1}}=\frac13\Big(1+\frac{a_{n-2}}{a_{n-1}}+a_{n-2}\Big)$$

Show that $a_n\xrightarrow{n\rightarrow\infty}0$ and deduce that $x_n:=\frac{a_n}{a_{n-1}}$ converges to some $x>0$ satisfying $$x=\frac{1}{3}\Big(1+\frac{1}{x}\Big)$$

Recall that $$\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}|}{|a_n|}$$

Another approach, is to consider the discrete dynamical system on $\mathbb{R}^2$ given as \begin{align} \boldsymbol{x}_{n+1}&=G(\boldsymbol{x}_n)\\ \boldsymbol{x}_0 &= \begin{pmatrix} 0\\1\end{pmatrix} \end{align} where $$ G(x,y)=\begin{pmatrix} 0 & 1\\ \frac13 & \frac13 \end{pmatrix} \begin{pmatrix} x\\ y \end{pmatrix} + \frac13\begin{pmatrix} 0\\ xy \end{pmatrix} $$ The eigenvalues of the linear part of $G$ around the critical point $\boldsymbol{x}^*=[0\;0]^\intercal$ are $\frac{1\pm\sqrt{13}}{6}$.

From this, one can show that with initial conditions closed enough to $\boldsymbol{x}^*$, $\boldsymbol{x}_n$ converges to $\boldsymbol{x^*}$ exponentially fast (the rate of convergence being $\frac{1+\sqrt{13}}{6}$ which is happened to be $\lim_n\frac{a_{n+1}}{a_n}$ where $\boldsymbol{x}_n:=[a_{n-1}\,a_n]^\intercal $.

Solution 2:

I finally figure out the details to follow Oliver's hint and I would like to write out an answer here.

The two essential steps are proving that the sequences $(a_n)$ and $(\frac{a_n}{a_{n-1}})$ are both convergent. These two steps are nontrivial (!) and they are answered in the following two follow-up posts:

• Convergence of a recursively defined sequence of complex numbers: $3z_{n}=z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}$
• Finding $\lim_{n\to\infty} \frac{z_{n}}{z_{n-1}}$ where $3z_{n}=z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}$

It then follows by straightforward calculations that $$ \lim_{n\to\infty}\frac{a_{n}}{a_{n-1}} = \frac16(1+\sqrt{13}). $$

Consequently, by a well-known theorem (see, e.g., Rudin's Principle of Mathematical Analysis) which compares the ratio and root test: $$ \liminf _{n} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|} \leq \liminf _{n} \sqrt[n]{\left|a_{n}\right|} \leq \lim _{n} \sup \sqrt[n]{\left|a_{n}\right|} \leq \limsup _{n} \frac{\left|a_{n+1}\right|}{\left|a_{n}\right|} $$ the desired result follows.

Remark.

It is worthing knowing that, in Python, one needs a good algorithm to calculate the $n$-th root for very big $n$. In the original attempt mentioned in the post, the total iteration number was taken too big that the plot is actually not accurate because a**(1/n) was used in the code to find the $n$-th root. However, if one only iterates 100 times, one gets $b_{100}\approx 0.76759187924+\pm 10^{-11}$ which is an approximate value of $\frac{1}{6}(1+\sqrt{13})$.