The number $x_n$ satisfy the recurrence relation $$ x_{n+1}=\frac{[1+x_n+x_{n}^{2}]}{[2+x_n]}. $$ If $x_1>-2$, show that $[x_n - x_{n+1}]$, $[x_{n}-1]$, $[x_{n+1}-1]$ all have the same sign when $n>1$.
If $x_1>-2$, show that $x_{n}$ tends to $1$ as $n$ tends to infinity.
I understand that sequence has to be evaluated over the real field name but for this question, I am having difficulties recognizing what I am suppose to do. Anyone to guide me or provide ideas if not to solve the question. If I were required to test the convergence of such a sequence I think I could have tried but this sounds so abstract to me. Thank you in advance mathematicians.
Well, first, look at the numerator. See that
$$\require{cancel}x_n>-2\implies1+x_n+x_n^2\ge\frac34>0$$
Similarly, the same is true for the denominator.
$$2+x_n>0$$
Since both the numerator and denominator must be positive, $x_2$ must be positive and
$$x_2\ge\frac12$$
To show that $x_n-1$ has the same sign for $n>1$, consider $x_1<-1$ or $x_1>1$, $x_2>1$. Thus, in those scenarios, we must show $x_n>1$. For $-1<x_1<1$, we must show $x_n<1$.
If $x_n>1$, $x_{n+1}>1$, and similarly in the other way around for $x_n<1$. You may prove by induction that for any $n>1$, $x_n$ will always have the same sign then.
By induction, this also proves $x_{n+1}-1$ will always have the same sign.
To show that $x_n-x_{n+1}$ will always have the same sign, use the recursive definition of $x_{n+1}$:
Assume it will always be positive:
$$\begin{align} x_n & >x_{n+1} \\ x_n & >\frac{1+x_n+x_n^2}{2+x_n} \\ \cancel2x_n+\cancel{x_n^2} & > 1+\cancel{x_n}+\cancel{x_n^2} \\ x_n & > 1 \end{align}$$
So $x_n>x_{n+1}$ if $x_n>1$. Similarly, one can show that $x_n<x_{n+1}$ if $x_n<1$ and $x_n=x_{n+1}$ if $x_n=1$.
For the sequence to converge, $\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n=L<\infty$.
$$\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}x_n=\lim_{n\to\infty}\frac{1+x_n+x_n^2}{2+x_n}$$
$$L=\frac{1+L+L^2}{2+L}$$
$$\cancel2L+\cancel{L^2}=1+\cancel L+\cancel{L^2}$$
$$L=1$$
Thus, if the limit exists, it is equal to $1$.
To show it converges, see that it is bounded and that from the fact that $a_n-a_{n+1}$ will always have the same sign, we must have
$$1<\dots<a_4<a_3<a_2\\ \text{or} \\ a_2<a_3<a_4<\dots<1$$
or the trivial $a_n=1$.
Since the sequence cannot diverge, it must converge, and if it converges, it converges to $1$.