Given that :
- $\ U_{0} =\frac{3}{4} $
- $\ U_{n+1} = \frac{U_{n}^2}{2U_{n}^2-2U_{n}+1}, n \in \mathbb{N}$
The question was to prove by recursion that : $\ \frac{1}{2}\lt U_{n} \lt 1$
I tried to solve this recursion, but at the end I found: $ \frac{1}{8} \lt U_{n+1} \lt 2$
- $\ \frac{1}{2}\lt U_{n} \lt 1$
- $\ \frac{1}{4}\lt U_{n}^2 \lt 1$
- $\ \frac{1}{2}\lt 2U_{n}^2 \lt 2$
- $\ -\frac{1}{2}\lt 2U_{n}^2-2U_{n} \lt 0$
- $\ \frac{1}{2}\lt 2U_{n}^2-2U_{n} +1 \lt 1$
- $\ {1}\gt \frac{1}{2U_{n}^2-2U_{n}+1} \gt \frac{1}{2} $
- $\ {2}\gt \frac{U_{n}^2}{2U_{n}^2-2U_{n}+1} \gt \frac{1}{8} $
- $\ {2}\gt U_{n+1} \gt \frac{1}{8} $
What am I doing wrong? or is this question wrong?
Let $f(x)=\frac{x^2}{2x^2-2x+1}.$ Then $$ f(x)=\frac{x}{2x^2-2x+1}=\frac{x^2}{x^2+(1-x)^2} =\frac{1}{1+(1/x-1)^2}. $$ On the interval $(\frac{1}{2},1),$ the function $(1/x-1)^2$ is decreasing, so $f(x)$ is increasing. Notice that $f(1/2)=1/2$ and $f(1)=1$. Therefore for $\frac{1}{2}<x<1$ we have $\frac{1}{2}<f(x)<1$. Base case: $\frac{1}{2}<U_0<1$. Induction step: if $\frac{1}{2}<U_n<1$ then $U_{n+1}=f(U_n)$ and so $\frac{1}{2}<U_{n+1}<1.$