Solving three alike problems $a_{n+1} = \frac{1+a_n^2}{1+a_n}$

Question

I have two problems alike:

1. If $a_1=a$, $a>0$, $a_{n+1} = \dfrac{1+a_n^2}{1+a_n}$, study its convergence.
2. If $a_0 >0$, $a_1>0$ and $a_{n+2} = \dfrac{2+a_{n+1}}{2+a_n}$, study this one's convergence.

I can solve this using two different situations: when $a$ is between $0$ and $1$ and when $a$ is between $1$ and $+\infty$ for the first problem and then say it is monotonic and it has an lower bond, therefore is convergent. But for me this solution does not help for the second, similar problem. Could you either show me how to solve the second one, not necessarily using my way, or show me a method that works for both? Thank you.

Answer 1

Too many questions. I'll take the 1st one.

Evidently $a_n>0\Rightarrow a_{n+1}=\frac{1+a_n^2}{1+a_n}>0$

Also $a_n>1\Rightarrow a_n^2>a_n\Rightarrow a_{n+1}=\frac{1+a_n^2}{1+a_n}>1$

and $a_n<1\Rightarrow a_n^2<a_n\Rightarrow a_{n+1}=\frac{1+a_n^2}{1+a_n}<1$

Now $a_{n+1}-a_{n}=\frac{1-a_n}{1+a_n}$

So if $a>1$ then $a_n$ is decreasing and bounded $a_n>1$ and if $a<1$ then $a_n$ is increasing and bounded $a_n<1$

In both cases it convergs, say , to $L$. By taking limits in both sides of $a_{n+1}=\frac{1+a_n^2}{1+a_n}$ we have $L=\frac{1+L^2}{1+L}\Rightarrow L=1$

The case $a=1$ is trivial : $a_n=1$

Answer 2

Hint with the 2nd one, it is easy to show by induction that $a_n>0$, then $$\color{red}{0<a_{n+2}}=\frac{2+a_{n+1}}{2+a_{n}}< \frac{2+a_{n+1}}{2}= 1+\frac{a_{n+1}}{2}< 1+\frac{1+\frac{a_{n}}{2}}{2}= 1+\frac{1}{2}+\frac{a_n}{2^2}<\\ 1+\frac{1}{2}+\frac{1}{2^2}+\frac{a_{n-1}}{2^3}<...<\\ 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{n}}+\frac{a_{1}}{2^{n+1}}\color{red}{< 2+\frac{a_{1}}{2^{n+1}}}$$ so $(a_n)$ is bounded. More to come later ...