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

calculusinductionsequences-and-series

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

Best Answer

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

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