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

Question

I have two problems alike:

If $a_1=a$, $a>0$, $a_{n+1} = \dfrac{1+a_n^2}{1+a_n}$, study its convergence.
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.

What do $x_0, x_1$ have to do with the second problem? Should that be $a_0, a_1?$ — saulspatz, Mar 04 '18 at 06:09
If you have three questions, you should ask three questions. They aren't any more similar than any two mathematical problems, non-linear recurrences are very rarely "similar". — , Mar 04 '18 at 08:15
I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. — GNUSupporter 8964民主女神地下教會, Mar 15 '18 at 16:43

score 1 · Answer 1 · answered Mar 04 '18 at 08:32

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 : $a_n=1$

score 0 · Answer 2 · answered Mar 04 '18 at 11:55

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 ...

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

2 Answers2