Suppose$a_1>0$,$a_2>0$,$a_{n+2}=\frac{2}{a_n+a_{n+1}}$.Prove that there exists $\lambda\in (0,1)$and a constant C,such that
$|a_n-1|<C\lambda^n$forall n $\in \mathbb{N} $.Furthermore,find the minimal $\lambda$ which suit $|a_n-1|<C\lambda^n$.
I can see that $\lim_{n\to \infty}a_n=1$.The proof below can be found in kaczor's book "Problems in Mathematical Analysis I".In fact,if $\frac1a\leqslant a_n,a_{n+1}\leqslant a$,then $\frac{1}{a}\leqslant a_{n+2}=\frac{2}{a_n+a_{n+1}}\leqslant a$.Thus ,by the principle of induction ,the sequence $\{a_n\}$is bounded.Put
$$l=\varliminf_{n\to \infty}a_n,L=\varlimsup_{n\to \infty}a_n$$
then for an arbitrarily fixed $\varepsilon>0$there exist $n_1,n_2\in\mathbb{N}$,such that
\begin{align*}
a_n<L+\varepsilon ,for \;n>n_1\tag{i}\\
a_n>l-\varepsilon ,for\; n>n_2\tag{ii}
\end{align*}
By (i),$a_{n+2}=\frac{2}{a_n+a_{n+1}}>\frac{1}{L+\varepsilon}$,$n>n_1$.Since the positive $\varepsilon$ can be arbitrarily small,we get $l\geqslant \frac1L$.In much the same way (ii) implies that $L\leqslant \frac1l$.Thus $l=\frac1L$.Let $\{n_k\}$be a sequence of positive integers such that $\lim_{k\to \infty}a_{n_k+2}=L$.We can assume that the sequences $\{a_{n_k+1}\}$,$\{a_{n_k}\}$and $\{a_{n_k-1}\}$converge to $l_1,l_2$and $l_3$,respectively.In fact ,if this is not the case,we can choose subsequences which do.By the definition of $\{a_n\}$,
$$l_1+l_2=\frac{2}{L}=2l,\;l_2+l_3=\frac{2}{l_1}$$
and since $l\leqslant l_1,l_2,l_3\leqslant L$,we get $l_1=l_2=l$and $l_2=l_3=L$.Hence $l=L$.This and the equality$l=\frac1L$imply that the sequence $\{a_n\}$converges to $1$.
It's hard for me to prove $|a_n-1|<C\lambda^n$,I've been thinking about this for one day, but I don't have any clue.
Best Answer
Let $b_n=a_n-1$. By writing the recurrence relation $a_{n+2}(a_n+a_{n+1})=2$ in terms of $b_n$, we find that $$ (1+b_{n+2})(b_n+b_{n+1}) + 2b_{n+2} = 0. $$
Now, as you have shown, $\lim_{n\to\infty}a_n=1$, or equivalently $\lim_{n\to\infty}b_n=0$. Thus the asymptotic rate at which $b_n$ tends to $0$ will be the same as that of $c_n$, where $c_n$ satisfies the recurrence relation $$ c_n+c_{n+1}+2c_{n+2}=0. $$ The characteristic equation of this recurrence is $2x^2+x+1=0$, both of whose solutions are complex with magnitude $1/\sqrt{2}$. Consequently, the minimal $\lambda$ is $1/\sqrt{2}$.
To be more precise, the argument I have just given can be transcribed into a rigorous proof that, for all $\lambda>1/\sqrt{2}$, there exists $C_\lambda>0$ such that $|a_n-1|<C_\lambda \lambda^n$ for all $n\in\mathbb N$. However, to go the extra step and show that there exists $C_{1/\sqrt{2}}$ such that $|a_n-1|<C_{1/\sqrt{2}}2^{-n/2}$ for all $n$ requires an extra idea.