How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$

Question

How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$?

I think to find the limit, first I need to prove that the sequence is convergent then it is easy to find the limit :

$L = \frac{L+2}{L+1} \to L = \sqrt{2}$ because the sequence $>0$

any hint to prove that the sequence is convergent?

thanks

Welcome John! Try using the fact that $a_{n+1} = 1 + 1/(a_n+1)$ to come up with some bounds. But yes, you are correct in proceeding to find the limit. — MrMazgari, Dec 12 '19 at 21:39
You can prove by induction that $a_{2n+1}$ is increasing, that $a_{2n}$ is decreasing and $a_{n}\geq 1$, and since $|a_{2n+1}-a_{2n}|=|\frac{1}{a_{2n}+1}-\frac{1}{a_{2n-1}+1}|<\frac{1}{4}|a_{2n-1}-a_{2n-2}|$ it converges. — egorovik, Dec 12 '19 at 21:44
Does this answer your question? Show that ${a_n}$ defined by $a_{n+1}=\frac{a_n+2}{a_n+1}$ converges — egorovik, Dec 12 '19 at 21:51

score 0 · Answer 1 · answered Dec 12 '19 at 21:57

0

Hint:

Compute $a_{n+1}-a_n$ in function of $a_n$ and $a_{n-1}$, and show that $$|a_{n+1}-a_n|<k\,|a_n-a_{n-1}| \quad (0<k<1),$$ then rewrite $a_n-a_0$ as $$a_n=(a_n-a_{n-1})+(a_{n-1}-a_{n-2})+\dots +(a_1-a_0).$$

answered Dec 12 '19 at 21:57

Bernard

179,256

score 0 · Answer 2 · answered Dec 12 '19 at 23:18

First, suppose the sequence converges to some $L$. We have

$$a_{n+1}=\frac{a_n+2}{a_n+1}=1+\frac{1}{1+a_n},$$ and we can take the limit of both sides. On the left,

$$\lim_{n\to\infty}(a_{n+1}-L+L)=\lim_{n\to\infty}(a_{n+1}-L)+L=L.$$

Now since $a_n>0$ (or even $a_n>1$) for all $n$, we know that $\lim\limits_{n\to\infty}a_n\neq -1$. Thus the limit of the right hand side is

$$\lim_{n\to\infty}\frac{a_n-L+L+2}{a_n-L+L+1}=\frac{\lim\limits_{n\to\infty}(a_n-L+L+2)}{\lim\limits_{n\to\infty}(a_n-L+L+1)}=\frac{L+2}{L+1}.$$ As you noted, this gives $L^2=2$. Of course $L\ge 0$, so $L=\sqrt 2$.

As you also noted, this is not a proof, since we began by assuming $L$ exists. We can show this sequence is Cauchy as follows. Write

\begin{align} a_{n}-a_{n-1}&=1+\frac{1}{1+a_{n-1}}-1-\frac{1}{1+a_{n-2}}\\ &=\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}. \end{align} In particular, $a_n>1$ for all $n$, and so

$$|a_{n}-a_{n-1}|=\left|\frac{a_{n-2}-a_{n-1}}{(1+a_{n-1})(1+a_{n-2})}\right|<\frac12|a_{n-1}-a_{n-2}|.$$

By induction,

\begin{align} |a_n-a_{n-1}|&<\frac12|a_{n-1}-a_{n-2}|\\ &<\frac{1}{2^2}|a_{n-2}-a_{n-3}|\\ &\hspace{3.5pt}\vdots\\ &<\frac{1}{2^{n-2}}|a_2-a_1|\\ &=\frac{1}{2^{n-1}} \end{align}

which shows the sequence is Cauchy.

score 0 · Answer 3 · answered Dec 13 '19 at 03:49

A standard trick is to take the fixpoint $L= \sqrt{2}$ of the equation and to consider

$a_n - L = a_n - \sqrt{2}$

Using

$a_{n+1} = 1+\frac{1}{a_n + 1} \stackrel{a_1 = 1}{\geq} 1 \Rightarrow a_n \geq 1$ for all $n\in\mathbb N$

you get immediately:

$$0\leq \left|a_{n+1} - \sqrt{2} \right| = \left|1+\frac{1}{a_n + 1} - \left(1+\frac{1}{\sqrt 2 + 1}\right) \right|$$ $$= \left|\frac{\sqrt 2 - a_n}{(a_n + 1)(\sqrt 2 + 1)}\right|\stackrel{a_n \geq 1}{\leq}\underbrace{\frac 1{2(\sqrt 2 + 1)}}_{= q < 1}\left|a_n - \sqrt 2\right|$$ $$\leq q^n\left|a_1 - \sqrt 2\right|= q^n(\sqrt 2-1)\stackrel{n\to\infty}{\longrightarrow}0$$

How to find the limit of $a_1=1\:,\:a_{n+1}=\frac{a_n+2}{a_n+1}$

3 Answers3