Convergence of recursive sequence $x_n=\sqrt{x_{n-1}^2-x_{n-1}+1}$ where $0<x_0<1$

Question

as you can see from the title I have a recursive sequence with the type above. I want to prove that this sequence converges. Although I have searched a lot for some related sequence, I haven't figured out how to prove that it is monotonous. Which is the preferred way of solving this?

Answer 1

Let $f(x)=\sqrt{x^2-x+1}$. The sequence can be written then as $x_n=f(x_{n-1})$.

On $(0,1)$ we have $x< f(x)<1$. From here it follows easily that $x_n>x_{n-1}$ and $x_n<1$ for all $n$.

Answer 2

It should read: $x_n=\sqrt{x_{n-1}^2-x_{n-1}+1}$ , where $ 0<x_0<1$.

1. Show by induction: $0<x_n <1$ for all $n$.

2. Show by induction: $(x_n)$ is increasing.

Answer 3

Ok, I think that I have a solution, first of all I showed that $0<x_1<1$ . Since $0<x_0<1$, this means that also $0<x_0^2<1$ and also that $x_0^2<x_0$ <=> $x_0^2-x_0<0$ <=> $x_0^2-x_0+1<1$, which means that $x_1=\sqrt{x_0^2-x_0+1}<1$. So in the same manner I proved that if $0<x_n<1$ then also $0<x_{n+1}<1$. After that I showed that $x_{n+1}-x_n>0$, so $x_n$ is increasing until it reaches 1.

Answer 4

I haven't figured out how to prove that it is monotonous

It is given that $0 \lt x_0 \lt 1\,$, and suppose by induction that $0 \lt x_{n-1} \lt 1\,$. Then:

• $$\require{cancel} 1 - x_n^2 = \cancel{1}-(x_{n-1}^2-x_{n-1}+\cancel{1})=x_{n-1}-x_{n-1}^2=x_{n-1}(1-x_{n-1}) \;\in\; (0,1) $$

• $$ x_n=\sqrt{x_{n-1}^2-2x_{n-1}+1+ x_{n-1}}=\sqrt{(x_{n-1}-1)^2+x_{n-1}} \gt \sqrt{x_{n-1}} \gt x_{n-1} $$

Therefore $0 \lt x_n \lt 1$ and $x_n \gt x_{n-1}\,$, which completes the induction step and proves both monotonicity and boundedness to $(0,1)\,$.