Prove or disprove: The recursive sequence $(a_{n})_{n=1}^\infty: a_{n+1} = -3 - a_{n}^2, a_1 = 2$ converges

Question

My intuition is that this sequence is not convergent. I started by proving with induction that the sequence is strictly decreasing:

Base case:

$n = 1, a_1 = 2$

$n = 2, a_2 = -3 - (2)^2 = -7 < a_1$

Induction step:

Assume $a_n < a_{n-1}$

$a_{n+1} = -3 - a_n^2 < -3 - a_{n-1}^2 < a_n$

Thus $(a_n)_{n=1}^\infty$ is strictly decreasing

Next I tried to arrive at a contradiction using the definition of the limit of a sequence

ATC that $a_n$ converges to some L $\in$ R

Let $\varepsilon$ > 0, N $\in$ N to be defined later such that $\forall$n $\geq$ N we have

| $a_n$ - L | < $\varepsilon$

L - $\varepsilon$ < $a_n$ < L + $\varepsilon$

At this point I'm unsure if this proof is going on the right direction or if I took the wrong approach

Answer 1

Assume that $a_n$ converges to some $L\in\mathbb{R}$. Then $L=-3-L^2$, but this equation hasn't real solutions.

Answer 2

Let $ n $ be a positive integer, we have : \begin{aligned}-\frac{a_{n+1}}{a_{n}}&=\frac{3}{a_{n}}+a_{n}\\ \iff \left(\frac{a_{n+1}}{a_{n}}\right)^{2}&=\frac{9}{a_{n}^{2}}+6+a_{n}^{2}\\ \frac{a_{n+1}^{2}}{a_{n}^{2}} &\geq6\\ \Longrightarrow\ \ \prod_{k=1}^{n-1}{\frac{a_{k+1}^{2}}{a_{k}^{2}}}&\geq\prod_{k=1}^{n-1}{6}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ a_{n}^{2}&\geq 4\times 6^{n-1}\\ \iff \ \ \ \ \ \ \ \ \ \left|a_{n}\right|&\geq 2\times \sqrt{6}^{n-1} \end{aligned}

Thus $ \lim\limits_{n\to +\infty}{\left|a_{n}\right|}=+\infty $, which means $ \left( a_{n}\right)_{n} $ is a divergent sequence.