$\lim x_0^2 + x_1^2+...+x_n^2$ where $x_n=x_{n-1}-x_{n-1}^2$

Question

So, we are given a sequence $x(n)$ for which $x_{n+1}= x_n-x_n^2$ , $x_0=a$, $0 \le a \le 1$

I was first requested to show that it converges and to find $\lim_{n \to \infty} x_n$. I will post my answer here for you to check if it is right :S

so, $x_n-x_{n+1} = x_n-x_n+x_n^2>0$ , thus the sequence is decreasing and since every term is positive its also bounded, therefore it converges to its infimum=A

so $A=\lim x_n= \lim x_{n+1}= \lim (x_n-x_n^2) = A-A^2 => A^2=0 => A=0$

And then it asks for me to find $\lim_{n \to \infty} x_0^2+x_1^2+...+x_n^2$ . I tried to do the same by defining a different sequence but it didn't work.

I know it converges but I can't find the limit. Can you give me any help? If it is possible, don't post any answer that involves series because I haven't studied them yet.

Please type properly, otherwise it is very difficult to read. — Samrat Mukhopadhyay, Jul 24 '13 at 18:20
@Plom: I edited some of your posting. It really helps readability to post using MathJax (see FAQ). More improvements can be made to the post, but I wanted to jump start you so you can see if you can improve. Regards — Amzoti, Jul 24 '13 at 18:22
i dont know how to use latex if thats what u mean, i ll try to get around with learning it though, sorry for now — Plom, Jul 24 '13 at 18:22
i dont know what telescopic is :S sorry if im being too much of a pain — Plom, Jul 24 '13 at 18:26

score 2 · Accepted Answer · answered Jul 24 '13 at 18:28

2

Notice that by rearranging the recursive formula, we obtain $x_n^2=x_n-x_{n+1}$. Plugging this in yields:

$$x_0^2+x_1^2+\ldots+x_n^2=(x_0-x_1)+(x_1-x_2)+\ldots+(x_n-x_{n+1})=x_0-x_{n+1}$$

Now taking the limit gives $x_0=a$ because $x_{n+1}$ vanishes as $n$ goes to infinity.

answered Jul 24 '13 at 18:28

Jared

31,451

omg, im so stupid -_- thx a bunch... – Plom Jul 24 '13 at 18:30

$\lim x_0^2 + x_1^2+...+x_n^2$ where $x_n=x_{n-1}-x_{n-1}^2$

1 Answers1