How to prove a sequence diverges? $a_n^\infty,$ defined by $a_{n+1}=-1-a_n^{2},\qquad a_1=1$

Question

So I'm given a sequence $a_n^∞$ that is defined in a recursive formula as follows: $$a_{n+1}=-1-a_n^{2},\qquad a_1=1$$ and they're asking prove/disprove that this sequence converges, so I'm thinking it diverges because as you plug in numbers you go further into negative values. The question is how to disprove it.. like I want to try contradiction but I don't know how to contradict this.

Every help is appreciated!

Answer 1

Let $b_n = -a_n$ for $n\ge 2$ and $b_1=1$.

Then $b_{n+1}=b^2_n+1 \ge 2b_n$, because $b_n^2-2b_n+1=(b_n-1)^2 \ge 0$.

Therefore, $b_n \ge 2^{n-1} b_1 = 2^{n-1}$ and so $a_n \le - 2^{n-1}$ and $a_n$ diverges to $-\infty$.

Answer 2

For convenience we can safely ignore the minus signs and consider

$$a_{n+1}=a^2_n+1.$$

It is obvious that

$$a_n\ge2\implies a_{n+1}=a_n^2+1>a^2_n\ge2a_n$$ so that the sequence is bounded below by a diverging geometric progression.

Answer 3

$a_1$ is a real number, and it's clear from the recurrence $a_{n+1} = -1 - a_n^2$ that $a_n$ must be real for all $n$. Hence, if $a_n$ converges, it must converge to a real limit.

Suppose that $a_n$ does converge, say to a limit $L$. Taking limits of both sides of $a_{n+1} = -1 - a_n^2$, we obtain $L = -1 - L^2$, or equivalently $L^2 + L + 1 = 0$. You can easily check that this equation has no real solutions (the $b^2 - 4ac$ term in the quadratic formula is $1 - 4 = -3$), so $L$ is not real. This contradiction implies that $a_n$ does not converge.

Answer 4

I think that the "best" answers are those already given, as they imply explicit bounds on $a_n$ that quantify the blow-up rate of the sequence, that is, the speed at which it diverges to infinity.

(EDIT What follows in the gray box is exactly the same as Bungo's answer)

However, there is another method of proof. Observe that the sequence $a_n$ is integer. Assume for a contradiction that $a_n\to x$. Since the sequence $a_n$ is integer, $x$ must be integer. Moreover, since $\lim_{n\to \infty} a_{n+1}=\lim_{n\to \infty} a_n$, we obtain for $x$ the equation $$ x^2+x+1=0.$$ Completing the square this equation can be rewritten as $(x+\tfrac12)^2+\tfrac34=0$ so it cannot have any real solution, let alone an integer one. (Thank you Daniel Wainfleet for the suggestion). We have reached the desired contradiction.

NOTE. We did not really use the assumption that $a_n$ is integer, we only used that it is real. It is therefore the case that the given recurrence relation $a_{n+1}=-1-a_n^2$ does not produce convergent sequences whatever the initial condition $a_0\in\mathbb R$ is.

It is interesting to compare with the recurrence relation $a_{n+1}=1 - a_n^2$. This leads to the equation $x^2+x-1=0$, with roots $x_\pm=\tfrac12(-1\pm\sqrt 5)$. These roots are not integer, and consequently the sequence $a_n$ cannot converge if $a_0$ is integer; to prove this, just apply the argument above.

However, this argument does not say anything in the case of noninteger $a_0$. This is no coincidence, for in this case there do exist noninteger initial data such that $a_n$ converges: taking $a_0=x_+$ or $a_0=x_-$, where $x_\pm$ are the roots found above, yields two constant sequences, hence convergent.

Changing a $+1$ to a $-1$ in the original recurrence produced a dramatic change in the behavior of the solutions.