## Task 6 - 261236

Let $\left(x_{n}\right)$ be the sequence of real numbers for which

$$
x_{1}=\sqrt{3} \quad(1) \quad \text { and } \quad x_{n+1}=\sqrt{9 x_{n}^{2}+11 x_{n}+3} \quad(n=1,2,3, \ldots)
$$

holds. For every real number $a \neq 0$, let $\left(y_{n}\right)$ be the sequence defined by

$$
y_{n}=\frac{x_{n}}{a^{n}} \quad(n=1,2,3, \ldots)
$$

Determine all $a \neq 0$ for which the sequence $\left(y_{n}\right)$ is convergent.
From (1) and (2) it follows by complete induction that $x_{n}>0$ for all $n=1,2,3, \ldots$ and then

$$
\begin{aligned}
& 9 x_{n}^{2} < 3^{n} / 2 \text{ for all } n=1,2,3, \ldots \text{ and then }
\end{aligned}
$$

$$
31$, so $\lim _{n \rightarrow \infty} q^{n}=+\infty$; on the other hand, it follows from (5) by complete induction $(n=1,2,3, \ldots)$

$$
\left|y_{n+1}\right|>q \cdot\left|y_{n}\right| \quad ; \quad\left|y_{n+1}\right|>q^{n} \cdot\left|y_{1}\right|
$$

and because $\left|y_{1}\right|>0$, it follows that $\lim _{n \rightarrow \infty}\left|y_{n}\right|=+\infty$. Therefore, the sequence $\left(y_{n}\right)$ is divergent in this case.

2. If $|a|>3$, then because

$$
\lim _{n \rightarrow \infty}\left(\frac{3}{|a|}+\frac{4}{|a| \cdot 3^{n}}\right)=\frac{3}{|a|}<1
$$

it follows that $\lim _{n \rightarrow \infty} q^{n}=0$. Therefore, the sequence $\left(y_{n}\right)$ is convergent in this case.

3. If $a=-3$, then $q=1$ and $y_{n+1}=-y_{n}$. Therefore, the sequence $\left(y_{n}\right)$ is divergent because it alternates between positive and negative values and does not approach a single limit.

4. If $a=3$, then $q=1$ and $y_{n+1}=y_{n}$. Therefore, the sequence $\left(y_{n}\right)$ is convergent because it is constant.

5. If $|a|<3$, then because

$$
\lim _{n \rightarrow \infty}\left(\frac{3}{|a|}+\frac{4}{|a| \cdot 3^{n}}\right)=\frac{3}{|a|}>1
$$

it follows that $\lim _{n \rightarrow \infty} q^{n}=+\infty$. Therefore, the sequence $\left(y_{n}\right)$ is divergent in this case.

Thus, it is shown: The sequence $\left(y_{n}\right)$ is convergent for all $a$ such that $a<-3$ or $a \geq 3$.

Adopted from $[5]$

# 9.28.4 IV. Round 1986, Class 12 

#### Abstract