Does the sequence $x_{n+1}=x_n+x_n^2$ converge to $0$ whenever $-1\lt x_0\lt0$?

Question

My question concerns using the fixed-point iteration to find the fixed point of the function $f(x)=x+x^2=x(1+x)$ (this function has a single fixed point at $0$).

The problem

Given some fixed $x_0$, define the sequence $x_n$ by $$\begin{align*} x_{n+1}=x_n+x_n^2&&\text{for all $n\geq0$.} \end{align*}$$ Does this sequence converge to $0$ for all $x_0$ in the range $-1\lt x_0\lt0$? After some computational numerical exploration, I think the answer might be yes but I'm not so sure how to prove this.

Some notes

The sequence trivially converges to $0$ when $x_0=-1$ or $x_0=0$. It's also fairly easy to prove that it does not converge to $0$ whenever $x_0$ lies outside this range (i.e., $x_0\lt-1$ or $x_0\gt0$).

I've already proved that for any $x_0$ in this range, all successive $x_n$s remain in this range (i.e., $-1\lt x_n\lt0$ for all $n\geq0$) and that the sequence increases / gets closer to $0$ (i.e., $x_{n+1}\gt x_n$), but is there any way to prove it actually converges to $0$?

Some examples

If, for example, $-\frac{1}{2}\lt x_n\lt-\frac{1}{4}$, we'll have $-\frac{3}{8}\lt x_{n+1}\lt-\frac{1}{8}$ and $-\frac{21}{64}\lt x_{n+2}\lt-\frac{5}{64}$, etc. So it seems as though the range of possible values gets closer to $0$. But, how would one prove this?

Answer 1

You can use the monotone convergence theorem. It tells us that a sequence $\{x_n\}_{n=1}^\infty\subseteq \mathbb R$ converges if it is bounded and decreasing. For you sequence we have $|x_{n+1}| = |x_n| |1+x_n| \leq 1$ since $x_0 \in (-1,0)$. And likewise we have $x_{n+1} = x_n(1+x_n) \leq x_n$ as $1+x_n \in (0,1)$.

(You can use induction to prove both these claims.)

Now you have shown that $\lim_{n\to\infty} x_n=x$ exists. From this we see that $$ x = \lim_{n\to\infty} x_{n+1} = \lim_{n\to\infty} x_n(1+x_n) = x(1+x). $$ Hence $x$ satisfies $x = x^2+x$ so $x=0$.

Answer 2

The answer to your question (regarding convergence to $0$) is indeed affirmative and for a reason that is not too difficult to ascertain. For simplicity let us introduce the quadratic polynomial function: \begin{align*} f \colon \mathbb{R} &\to \mathbb{R} \\ f(x)&=x^2+x \end{align*} and let us note that $f[(-1, 0)] \subseteq (-1, 0)$, in other words the interval $(-1, 0)$ is stable under $f$. To prove this latter claim it suffices to notice that $x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4} \geqslant \frac{3}{4}>0$ for any $x \in \mathbb{R}$ (and thus in particular $x^2+x>-1$ always) and respectively that given $x \in (-1, 0)$ we have $x+1>0$ yet $x<0$ which means that $x^2+x=x(x+1)<0$.

The immediate conclusion that we draw from this is that for any fixed $t \in (-1, 0)$, the unique recursive sequence $x \in \mathbb{R}^{\mathbb{N}}$ defined by $x_0=t$ and $x_{n+1}=f(x_n)$ for any $n \in \mathbb{N}$ has the property that actually $x \in (-1, 0)^{\mathbb{N}}$ and is thus a bounded sequence. Furthermore, since $x_{n+1}=x_n+x_n^2 \geqslant x_n$ takes place for every $n \in \mathbb{N}$ we gather that $x$ is increasing (one can actually make the more precise observation that since $x_n \neq 0$ for all $n \in \mathbb{N}$ under the stated conditions, the inequality mentioned above is actually strict, rendering $x$ into a strictly increasing sequence, however that is not essential for the study of convergence).

According to one of the theorems of Weierstrass, any upper-bounded increasing sequence of real numbers is convergent, which applies in particular to $x$. Since $x$ is defined recursively with respect to the continuous function $f$, it follows with necessity that $\displaystyle\lim_{n \to \infty} x_n$ must be a fixed point for $f$. However it is obvious that there is just one such fixed point - the unique solution to the equation $f(u)=u \Leftrightarrow u^2=0$ - which is $0$.