TITLE: How show that $|a_{n}-1|\le c\lambda ^n,\lambda\in (0,1)$
QUESTION [15 upvotes]: I'm looking for proof that
Let $a_{0},a_{1}>0$,and such
$$a_{n+2}=\dfrac{2}{a_{n+1}+a_{n}}$$
show that: there exist $\lambda,c>0$ such
$$|a_{n}-1|\le c\lambda ^n,\lambda\in (0,1)$$
I tried using induction, but i'm not sure how it would work ,Thanks in advance.

REPLY [2 votes]: Here's a short elementary solution which does not involve multivariable calculus or linearisation arguments.

Motivation: The size of any two consecutive terms limit the size of all the terms that follow it, in the sense that for any $M\geq1$, $a_n,a_{n+1}\in[\frac1M,M]$ implies $a_k\in[\frac1M,M]$ for all $k\geq n$. We expect that the value of this $M$ cannot stay large, since eg. if $a_n,a_{n+1}$ are very large, then $a_{n+2}$ is very small, and hence $a_{n+3},a_{n+4}$ will be more 'moderate' in size.
  Thus we want to prove a bound of the following form: if $a_n,a_{n+1}\in[\frac1M,M]$ then $a_{n+3},a_{n+4}\in[\frac1{M'},M']$, where $M'-1\leq C(M-1)$ for some $C<1$. This is enough to show the exponential decay of $|a_n-1|$.

Define $M_n=\sup_{k\geq n}(a_k,\frac1{a_k})$; note that $M_n\geq1$, and $(M_n)_{n\geq0}$ is nonincreasing.
I claim that $(M_{n+3}-1)\leq\frac34(M_n-1)$ for all $n\geq0$; if this is true then there exists $c$ such that $M_n-1<c\left(\sqrt[3]{\frac34}\right)^k$, and we are done because
$$1-(M_n-1)<\frac1{M_n}\leq a_n\leq M_n=1+(M_n-1).$$
Here we split into 4 easy cases. From here on, write $M=M_n$ for convenience.
Case 1: $a_n,a_{n+1}\in[1,M]$. Then we can get
\begin{align*}
a_{n+2}&\in[\frac1M,1],\\
a_{n+3}&\in[\frac2{1+M},\frac{2M}{1+M}],\\
a_{n+4}&\in[\frac{2(1+M)}{1+3M},\frac{2M(1+M)}{1+3M}].
\end{align*}
Now it is easy to check that $\frac2{1+M},\frac{2(1+M)}{1+3M}\geq\frac4{1+3M}$, and $\frac{2M}{1+M},\frac{2M(1+M)}{1+3M}\leq\frac{1+3M}4$. Hence $a_{n+3},a_{n+4}\in[\frac4{1+3M},\frac{1+3M}4]$, so $M_{n+3}\leq\frac{1+3M}4$, which is equivalent to the original claim.
Case 2: $a_n,a_{n+1}\in[\frac1M,1]$. Similarly,
\begin{align*}
a_{n+2}&\in[1,M],\\
a_{n+3}&\in[\frac2{1+M},\frac{2M}{1+M}],\\
a_{n+4}&\in[\frac{2(1+M)}{M(3+M)},\frac{2(1+M)}{3+M}].
\end{align*}
Analogously, we check that $\frac2{1+M},\frac{2(1+M)}{M(3+M)}\geq\frac4{1+3M}$, and $\frac{2M}{1+M},\frac{2(1+M)}{3+M}\leq\frac{1+3M}4$, which yields the same conclusion.
Case 3: $a_n\in[1,M]$, $a_{n+1}\in[\frac1M,1]$. Note that this is Case 1 with the index shifted by 1, so $M_{n+3}\leq M_{n+2}\leq\frac{1+3M}4$.
Case 4: $a_n\in[\frac1M,1]$, $a_{n+1}\in[1,M]$. Analogous to Case 3.