<s>[INST] Examining the convergence with parameter $a$ For $a \in R$, let $x_1=a$ and $x_{n+1}=\frac{1}{4}(x_{n}^2+3)$ for all  $n≥2$. Examine the convergence of the sequence ${x_n}$ for different values of $a$. Also, find $\lim x_n$, whenever it exists.
I am having problems on how to take $a$ as a parameter. I am unable to think which all values of $a$ would give different results.
Hints are appreciated. Thanks.
 [/INST] *

*Let $a>3$.


Hence, $$x_{n+1}-3=\frac{(x_n-3)(x_n+3)}{4}>0$$ and by induction we obtain here $x_n>3$ for all $n\in\mathbb N$. 
In another hand, $$x_{n+1}-3=\frac{(x_{n-1}-3)(x_{n-1}+3)}{4}>$$
$$>\frac{3}{2}(x_{n-1}-3)>\left(\frac{3}{2}\right)^2(x_{n-2}-3)>...>\left(\frac{3}{2}\right)^{n-1}(x_1-3),$$
which says that for $a>3$ our sequence does not converge. 


*$a=3$.


In this case $x_n=3$ and our sequence converges.


*$1<a<3$.
$$x_{n+1}-1=\frac{(x_{n}-1)(x_{n}+1)}{4},$$ which says that $1<x_n<3$ for all $n$ by induction and
$$x_{n+1}-x_n=\frac{(x_n-3)(x_n-1)}{4}<0,$$ which says that $\{x_n\}$ decreasing. 


Thus, there is $\lim\limits_{n\rightarrow+\infty}x_n$ and let this limit equal to $x$.
Thus, $1\leq x<3$ and $x=\frac{1}{4}(x^2+3),$ which gives $x=1$.


*$a=1$.


In this case $x_n=1$ and our sequence converges.


*$-3<a<1$.


Here $|x_n+1|\leq|a-1|<4$ for all $n$ and $$|x_{n}-1|=|\frac{(x_{n-1}-1)(x_{n-1}+1)}{4}|\leq|x_{n-1}-1|\cdot|\frac{a-1}{4}|\leq$$
$$\leq|x_{n-2}-1|\cdot|\frac{a-1}{4}|^2\leq|a-1|\cdot|\frac{a-1}{4}|^{n-1},$$
which says that $\lim\limits_{n\rightarrow+\infty}x_n=1$ by the limit definition.


*$a=-3$. 


In this case $x_2=3$ and $x_n=3$ for all $n\geq2$, which says that $x$ converges.


*$a<-3$.


Since $x_{n+1}-3=\frac{(x_n-3)(x_n+3)}{4}>0$ by induction and $x_{n+1}-x_n=\frac{(x_n-3)(x_n-1)}{4}>0$,
we see that our sequence has no a limit.
Done!
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