Sequence convergence with parameter

Question

I need to find the parameter $a$ so that the following series converges: $ x_0=a$, $x_{n+1}=x_n^2-4x_n+6$.What is the condition for this series to converge and why?

Answer 1

Since $x_{n+1}-2=\left(x_n-2\right)^2$, we can prove by induction that $$x_n=(a-2)^{2^n}+2$$ and the rest is smooth.

Answer 2

We can consider the equation: $$a^2 - 4a + 6 = a$$ Which has solutions $a=2$ and $a=3$ (If the sequence were to converge, then it would converge either to $2$ or $3$ , depending on $a$). Additionally, we can check that, for $a>3$: $$a^2-4a+6 > a >3$$ So for these values, the sequence will not converge. Now, if $a<1$ We can also check that: $$a^2-4a+6 > 3 \iff (a-3)(a-1)>0$$ So if $a>1$, then $x_1 >3$ and $x_n$ diverges. Lastly, for $a \in [1,3]$ if $a \in (2,3)$, then $x_n$ is decreasing, but bounded by $2$, therefore it goes to $2$. If $a \in (1,2) $, then $x_1 \in (2,3)$, and $x_n$ again converges to $2$. Finally, if $a=1,2,3$, then $x_n$ converges to $3,2,3$ respectively because it becomes constant. To summarize: $$a\in(-\infty,1) \cup (3,+\infty) \iff x_n \ \ \ \text{diverges}$$