Prove that the sequence defined by a recurrence is bounded

Problem. Let, $$a(1)=1,\quad a(2)=e,\quad a(n+1)=\sqrt[3]{a^{2}(n)a(n-1)},\quad \forall n\geqslant 2.$$ Prove that $(a_{n})_{n=1}^{+\infty}$ is not convergent.

EDIT: @belows Thanks for the corrections, it looks like the sequence is convergent.

My approach:

Calculating the first terms in the sequence we can see that

$$a(1)=1,\quad a(2)=e,\quad a(3)=e^{2/3},\quad a(4)=e^{7/9}, \ldots$$ so in particular we have $$a(2)>a(3)<a(4)$$ Thus $(a_{n})$ is not monotocally increasing and decreasing, that means, the sequence is not monotocally.

It's clear that $0<a_{n}$ for all $n\in \mathbb{N}$ so $(a_{n})_{n}$ is bounded below.

But, how can I prove that $(a_{n})_{n}$ is bounded above? I think we can see that for all $n\in \mathbb{N}$, $a_{n}<3$ but I don't sure about this.

Solution 1:

Hint: (For bounded.) If $1\leq b,c\leq e$ then show $1\leq \sqrt[3]{b^2c}\leq e.$

The series is convergent. Show $x_n=\log a_n$ satisfies $$|x_{n+1}-x_n|= \frac13|x_n-x_{n-1}|.$$ Show this means $(x_n)$ is Cauchy, thus convergent.

An alternate approach, if the logarithm i seems too magical, is to let $$y_n=\frac{a_{n+1}}{a_n}$$ and you get $$y_{n+1}=y_n^{-1/3}=\cdots=y_1^{(-1)^n/3^n}$$ Since $y_1=e, a_1=1,$ $$\begin{align} a_{n}&=a_1\cdot\prod_{k=1}^{n-1}a_k\\&=\prod_{k=1}^{n-1}e^{(-1/3)^{k-1}}\\&=e^{\sum_{k=1}^{n-1}(-1/3)^{k-1}} \end{align} $$

But the exponent is a geometric sequence, which converges.