Recurrence relation $x_{n+1}=x_n-x_{n}^2$ with $x_1=1/2$

Question

Given the recurrence relation $x_{n+1}=x_n-x_{n}^2$ with $x_1=1/2$, evaluate $$\lim_{n \to \infty} x_n$$ And $$\lim_{n \to \infty} nx_n$$ My attempt is this: $x_2=1/4$, $x_3=3/16$, $x_4=39/256$; the sequence seems to be decreasing, and by the recurrence I notice that $x_{n+1}=x_n-x_n^2 \leq x_n$ so it is decreasing.

Then it has a limit $L$ and by the recurrence it must be $L=L-L^2\Leftrightarrow L=0$; hence $$\lim_{n \to \infty} x_n =L=0$$ Don't know what to do for the second limit, because just evaluating it gives an indeterminate form $\infty\cdot0$.

Thanks.

Answer 1

Consider the inverse sequence $y_n=1/x_n$. We can write the following: $$ y_{n+1} = \frac1{x_{n+1}} = \frac1{x_n(1-x_n)} = \frac{y_n^2}{y_n-1} = y_n+\frac{y_n}{y_n-1}. $$

Can you see how fast this thing is growing when $n\gg0$?

Answer 2

hint

At first, the sequence is decreasing since

$$x_{n+1}-x_n = -x_n^2\le 0$$

secondly, it is easy to prove by induction that

$$(\forall n\in \Bbb N)\;\;\; 0\le x_n \le 1$$

decreasing and bounded , it converges to $ L$.

$[0,1]$ is closed thus $L\in[0,1]$.

and $L$ satifies $$L=L-L^2$$ because $ f : x\mapsto x-x^2$ is continuous at $ [0,1]$ and $$x_{n+1}=f(x_n).$$

finally, $L=0$.