Given that U_0=3/2
$U_{n+1}=U_n^2-2U_n+2$
Show that $1<U_n<=3/2$
I did it by induction
Then we have to show that
$U_{n+1}-1<1/2(U_n-1)$
And deduce that $U_n-1<=(1/2)^{n+1}$
Then we search values of n fir wich the number 1 will be an approchate value to $Un$ by $10^-6$
For seconde one i think i shpuld use induction
$U_{n+1}-1<=1/2(U_n-1)<=(1/2)^{n+1}<=(1/2)^{n+2}$
Then we deduce the sum U_0+U_1+...+U_n<=n+2-(1/2)^{n+1}
We have $$U_{n+1} = (U_n-1)^2+1 \implies U_{n+1}-1 = (U_n-1)^2 \implies U_{n+1}-1 = (U_n-1)(U_n-1)$$ Hence, it suffice to prove that $U_n \in [1,3/2]$. This is trivial from induction, since if $U_n \in [1,3/2]$, we have $(U_n-1)^2 \in [0,1/4] \implies U_{n+1} \in [1,5/4] \subseteq [1,3/2]$.