How show that $|a_{n}-1|\le c\lambda ^n,\lambda\in (0,1)$

Question

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.

Answer 1

First of all, we will prove that $a_n \to 1$.

Step 1. $(a_n)_{n \geqslant 0}$ is bounded.

Proof. We are looking for $m,M$ such that $m \leqslant a_n \leqslant M$ for every $n \geqslant 0$. If it's true for $n,n+1$ then : $$\frac{1}{M}\leqslant \frac{2}{a_n+a_{n+1}} \leqslant \frac{1}{m}$$ thus we need $mM=1$. Then we just chose $m$ wisely so that $m \leqslant a_0,a_1$.

Since the sequence is bounded, we set $\ell$ the inferior limit and $L$ the superior one.

Step 2. $\ell L=1$

Proof. Let $\phi$ be an extraction such that $a_{\phi (n)} \to \ell$ and by successive extractions we can assume that $a_{\phi (n)-1)} \to \ell _1 \in [\ell, L]$ and $a_{\phi (n)-2} \to \ell _2 \in [\ell, L]$. We obtain : $$\ell = \frac{2}{\ell _1 + \ell _2} \geqslant \frac{1}{L}$$ and then $\ell L \geqslant 1$. By replacing $\ell$ with $L$ we have $\ell L=1$

Step 3. $\ell = L=1$

Proof. It is the same idea, we set $\phi$ an extraction such that $a_{\phi (n)} \to L$, $a_{\phi (n)-1} \to \ell _1 \in [\ell , L]$, $a_{\phi (n)-2} \to \ell _2 \in [\ell , L]$, $a_{\phi (n)-3} \to \ell _3 \in [\ell , L]$. Then we have : $$\ell _1 + \ell _2 = \frac{2}{L}=2\ell $$ and $$\ell _2+ \ell _3 = \frac{2}{\ell _1}$$ Since $\ell _1, \ell_2 \geqslant \ell$ we obtain $\ell _1 = \ell _2 = \ell$ and then $\ell_2 + \ell _3= \frac{2}{\ell}=2L$ then $\ell_2=\ell_3 =L$ and we conclude $\ell =L$ : the sequence converges to $1$. Then let $N \geqslant 0$ such that for $n \geqslant N$ we have $|a_n+a_{n+1}| \geqslant 2- \varepsilon $

For your question.

let $U_n=(a_n,a_{n+1})$ so that $U_{n+1}=f(U_n)$ where $f(x,y)= \left(y,\frac{2}{x+y} \right)$. We can see that : $$A:= \mathrm{d} f(1,1)= \begin{pmatrix} 0 & 1 \\ -\frac{1}{2} & -\frac{1}{2} \end{pmatrix} $$

Fact. Maple or direct computation shows that $\displaystyle \rho (A) := \max \left \{ |\lambda|, \lambda \in \operatorname{sp} ( A) \right \} < 1 $

A classical lemma states that we can find a norm $\| \cdot \|_A$ such that : $$\| \mathrm{d}f (1,1) \|_A \leqslant \rho (A) + \mu < 1 $$ for a wisely chosen $\mu >0$. Then we set a ball $B:=B((1,1), \varepsilon$ such that $\|df(x)\|_A \leqslant \rho_0 <1 $ for every $x \in B$ (because $x \mapsto \mathrm{d}f(x)$ is continuous) and by the mean value theorem we obtain : \begin{align} \left \|\begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right \|_A & = \left \| f^{(n)} \left( \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} \right) - f^{(n)} \left( \begin{pmatrix} 1 \\ 1 \end{pmatrix}\right) \right \|_A \\ & \leqslant \left( \sup_{x \in B} \|\mathrm{d}f (x)\|_A \right)^{n-N} \left \|\begin{pmatrix} a_N \\ a_{N+1} \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right \|_A \end{align}

for $n \geqslant N$ where $N$ is chosen such that $a_n \in B$ for $n \geqslant N$.By norm equivalence in finite dimensional spaces there is a constant $C_1$ such that : \begin{align} |a_n-1| & \leqslant \max \left \{ |a_n-1|,|a_{n+1}-1| \right \} \\ & \leqslant C_1 \left \| \begin{pmatrix} a_n \\ a_{n+1} \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right \|_A \end{align}

Thus we have $|a_n-1| \leqslant c\lambda ^n $ for $n \geqslant N$ with : $$c= \sup_{x \in B} \|\mathrm{d}f (x)\|_A <1$$ and $$C=C_1 c^{-N} \left \|\begin{pmatrix} a_N \\ a_{N+1} \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \end{pmatrix} \right \|_A$$

Since we can let $C$ grow we can assume that the inequality holds for every $n \geqslant 0$.

Let $x_n := a_n - 1 \to 0$, and $X_n = (x_n,x_{n+1})$. We set $e_n := \|X_n\|$

The previous stuff proves that $\displaystyle e_n = \cal{O} \left( a^{ \frac{n}{2}}\right)$ for every $a> \dfrac{1}{2}$.

A simple computation and the fact that $x_nx_{n+1} \leqslant \frac{x_n^2 + x_{n+1}^2}{2}$ shows that : $$X_{n+1} = AX_n + {\cal{O}} \left( \| X_n \|^2 \right)$$

We set : $h_n:=2^{-n/2}e_n$. We can find a norm that we will still write $\| \cdot \|$ so that the associate matrix norm satisfies $\|A\| = \dfrac{1}{\sqrt{2}}$. A simple computation shows that : $$h_{n+1} \leqslant h_n + {\cal O} \left( \frac{h_n^2}{2^{n/2}} \right)$$ (I've used that $\|A X_n\| \leqslant \|A\| \|X_n\| \leqslant \dfrac{\|X_n\|}{\sqrt{2}}$ )

Thus $ \displaystyle h_{n+1} - h_n \leqslant {\cal O} \left( (1- \varepsilon)^n \right) $ for a wisely chosen $a$. Then $\sum h_{n+1} - h_n$ converges so that $h_n$ converge and we have $e_n \sim C 2^{-n/2}$.

I will investigate if we can go further, but it seems that we can, just setting $Y_n:=A^{n}X_n$ and study this.

I hope it is clear, and please feel free to correct my so weak english.

Answer 2

I think the idea is to linearize the sequence around $1$. Consider $e_n=a_n-1$. We know that the sequence $(e_n)$ converges to zero, and from the recurrence defining $e_n$ we have

$$e_n +e_{n+1}=\frac{2}{1+e_{n+2}}-2=-\frac{2e_{n+2}}{1+e_{n+2}} \tag1$$ Thus $$e_n +e_{n+1}+2e_{n+2}={\cal O}(e^2_{n+2})$$ Neglecting terms of higher order (linearizing), yields the linear recurrence $$e_n +e_{n+1}+2e_{n+2}=0$$ The characteristic equation of this linear recurrence sequence is $2X^2+X+1=0$ and it has two complex conjugate roots each of module $1/\sqrt2$. This proves that asymptotically $\vert e_n\vert={\cal O}(2^{-n/2})$. And proves the result.

Remark. In fact, $e_n\approx A 2^{-n/2} \cos(n\theta +\varphi)$, where $\theta=\arccos(-\sqrt2/4)$. So, I do not think that the sequence $(e_n)$ eventually alternate signs.

Answer 3

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.