Is it possible to utilize the convergence of the sequence $z_{n+1}=a/(1+z_n)$ to prove that the sequence $x_{n+2} = \sqrt{x_{n+1} x_n}$ is convergent?

Question

I'm doing Problem II.4.6 in textbook Analysis I by Amann/Escher.

For $x_0,x_1 \in \mathbb R^+$, the sequence $(x_n)_{n \in \mathbb N}$ defined recursively by $x_{n+2} = \sqrt{x_{n+1} x_n}$ is convergent.

My questions:

1. I'm not sure if my attempt (the parts from Lemma 4 to the end) is fine or contains logical gaps/errors. Could you please verify these parts? Any suggestion is greatly appreciated.

Particularly, I'm not sure if my induction in case $m > n$ in the part "...$\color{blue}{\text{vacuously true}}$ ..." and the proof that there exists $0 < \beta < 1$ such that $y_{n+1} \le \beta y_n$ are correct or not.

2. There is Problem II.4.5 as follows:

For $z_0,a \in \mathbb R^+$ , the sequence $(z_n)_{n \in \mathbb N}$ defined recursively by $z_{n+1}=a/(1+z_n)$ is convergent.

Using Mathematica, I found that both of the sequences $(x_n)_{n \in \mathbb N}$ and $(z_n)_{n \in \mathbb N}$ share the same plot as follows.

I would like to ask whether it is possible to utilize the convergence of $(z_n)_{n \in \mathbb N}$ to prove the convergence of $(x_n)_{n \in \mathbb N}$.

Thank you so much for your help!

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My attempt:

First we consider the case $x_0 < x_1$.

Lemma 1: $x_{2n} < x_{2n+1}$ for all $n$.

Proof: The statement trivially holds for $n=0$. Let it hold for some $n$. We have $$\begin{aligned} x_{2(n+1)} < x_{2(n+1)+1} & \iff x_{2n+2} < x_{2n+3} \\ &\iff \sqrt{x_{2n+1} x_{2n}} < \sqrt{x_{2n+2} x_{2n+1}} \\ &\iff x_{2n} < x_{2n+2} \\ &\iff x_{2n} < \sqrt{x_{2n+1} x_{2n}} \\&\iff x_{2n} < x_{2n+1}\quad (\star) \end{aligned}$$ in which $(\star)$ follows from inductive hypothesis. As such, the statement holds for $n+1$.

Lemma 2: $x_{2n} < x_{2n+2}$ for all $n$.

Proof: We have $x_{2n} < x_{2n+2} \iff x_{2n} < \sqrt{x_{2n+1} x_{2n}} \iff x_{2n} < x_{2n+1}$, which is true by Lemma 1. As a consequence, $(x_{2n})_{n \in \mathbb N}$ is increasing.

Lemma 3: $x_{2n+3} < x_{2n+1}$ for all $n$.

Proof: We have $x_{2n+3} < x_{2n+1} \iff \sqrt{x_{2n+2} x_{2n+1}} < x_{2n+1} \iff x_{2n+2} < x_{2n+1} \iff$ $\sqrt{x_{2n+1} x_{2n}} < x_{2n+1} \iff x_{2n} < x_{2n+1}$, which is true by Lemma 1. As a consequence, $(x_{2n+1})_{n \in \mathbb N}$ is decreasing.

Lemma 4: $x_{2m} < x_{2n+1}$ for all $m,n$.

Proof: In case $m \le n$, we have $x_{2m} \overset{(\star)}{\le} x_{2n} \overset{(\star\star)}{<} x_{2n+1}$ in which $(\star)$ follows from the fact that $(x_{2n})_{n \in \mathbb N}$ is increasing, and $(\star\star)$ follows from Lemma 1.

We prove the statement in case $m > n$ by induction on $m$. It's $\color{blue}{\text{vacuously true}}$ for $m=0$. Let it hold for some $m$. We have $$\begin{aligned} x_{2(m+1)} < x_{2n+1} & \iff x_{2m+2} < x_{2n+1} \\ &\iff \sqrt{x_{2m+1} x_{2m}} < x_{2n+1} \\ &\iff x_{2m+1} x_{2m} < x^2_{2n+1} \quad (\star) \end{aligned}$$ in which $(\star)$ follows from $x_{2m} < x_{2n+1}$ (by inductive hypothesis) and from $x_{2m+1} < x_{2n+1}$ (by $m > n$ and $(x_{2n+1})_{n \in \mathbb N}$ is decreasing). As such, the statement holds for $n+1$.

We define the sequence $(y_n)$ by $y_n := x_{2n+1} - x_{2n}$. We next prove that there exists $0 < \beta < 1$ such that $y_{n+1} \le \beta y_n$ for all $n$.

$$\begin{aligned} y_{n+1} < \beta y_n &\iff x_{2(n+1)+1} - x_{2(n+1)} < \beta (x_{2n+1} - x_{2n}) \\ &\iff x_{2n+3} - x_{2n+2} < \beta (x_{2n+1} - x_{2n}) \\&\iff \sqrt{x_{2n+2} x_{2n+1}} - x_{2n+2} < \beta (x_{2n+1} - x_{2n}) \\ &\iff \sqrt{x_{2n+2}} (\sqrt{x_{2n+1}} - \sqrt{x_{2n+2}}) < \beta (x_{2n+1} - x_{2n})\end{aligned}$$

Since $x_{2n+2} > x_{2n}$, $\sqrt{x_{2n+1}} - \sqrt{x_{2n+2}} < \sqrt{x_{2n+1}} - \sqrt{x_{2n}}$. As such, it suffices to prove that there exists $0 < \beta < 1$ such that $\sqrt{x_{2n+2}} (\sqrt{x_{2n+1}} - \sqrt{x_{2n}}) < \beta (x_{2n+1} - x_{2n})$. We have

$$\begin{aligned} &\sqrt{x_{2n+2}} (\sqrt{x_{2n+1}} - \sqrt{x_{2n}}) < \beta (x_{2n+1} - x_{2n}) \\ &\iff \sqrt{x_{2n+2}} < \beta (\sqrt{x_{2n+1}} + \sqrt{x_{2n}}) \\ &\iff \dfrac{\sqrt{x_{2n+2}}}{\sqrt{x_{2n+1}} + \sqrt{x_{2n}}} < \beta \\&\iff \left( \dfrac{\sqrt{x_{2n+2}}}{\sqrt{x_{2n+1}} + \sqrt{x_{2n}}}\right)^2 < \beta^2 \\ &\iff \dfrac{x_{2n+2}}{x_{2n+1} + 2\sqrt{x_{2n+1} x_{2n}} + x_{2n}} < \beta^2 \\ &\iff \dfrac{x_{2n+2}}{x_{2n+1} + 2x_{2n+2} + x_{2n}} < \beta^2\\ &\iff \dfrac{1}{2+ x_{2n+1}/x_{2n+2} + x_{2n}/x_{2n+2}} < \beta^2 \end{aligned}$$

As a result, we are done if we choose $1/\sqrt{2} <\beta < 1$. Then $y_{n+1} \le \beta y_n$ and thus $y_{n} \le \beta^n y_0$ for all $n$. We have $0 \le \lim_{n \to \infty} y_n \le \lim_{n \to \infty} \beta^n y_0 = 0$. As such, $\lim_{n \to \infty} y_n = 0$ and so $\lim_{n \to \infty}x_{2n} = \lim_{n \to \infty}x_{2n+1} = \alpha$.

From Lemmas 2,3, and 4, our sequence $(x_n)_{n \in \mathbb N}$ looks like $$x_0 < x_2 < x_4 < \cdots < x_{2n}< \cdots <x_{2n+1} < \cdots <x_5<x_3<x_1$$

By Nested Intervals Theorem, we have $$\lim_{n \to \infty}x_{2n} = \lim_{n \to \infty}x_{2n+1}$$

Next we prove that $$\lim_{n \to \infty}x_{n} = \alpha$$

Approach 1: For $\varepsilon > 0$, there exists $N \in \mathbb N$ such that $|x_{2n} - \alpha| < \varepsilon$ and $|x_{2n+1} - \alpha| < \varepsilon$ for all $n > N$. Thus $|x_{n} - \alpha| < \varepsilon$ for all $n > 2N$. As a result, $\lim_{n \to \infty}x_{n} = \alpha$.

Approach 2:

Given $n \in \mathbb N$, we have $A := \{2k+1 \in \mathbb N \mid k \ge n\} \subseteq B := \{k \in \mathbb N \mid k \ge n\}$ and, for each $k \in B$, there exists $k' \in A$ such that $x_k \le x_{k'}$. As such, $\sup_{k \ge n} x_{k} = \sup_{k \ge n} x_{2k+1}$ and thus $\inf_{n \ge 0} \sup_{k \ge n} x_{k} = \inf_{n \ge 0} \sup_{k \ge n} x_{2k+1}$. Similarly, we have $\sup_{n \ge 0} \inf_{k \ge n} x_{2k} =$ $\sup_{n \ge 0} \inf_{k \ge n} x_{k}$. It follows that $$\alpha = \sup_{n \ge 0} \inf_{k \ge n} x_{2k} = \sup_{n \ge 0} \inf_{k \ge n} x_{k} \le \inf_{n \ge 0} \sup_{k \ge n} x_{k} = \inf_{n \ge 0} \sup_{k \ge n} x_{2k+1} = \alpha$$ and thus $\lim_{n \to \infty} x_{n} = \alpha$.

The case $x_0 > x_1$ is similar, while the case $x_0 = x_1$ is trivial.

Answer 1

Your version is quite long. Here is a hint for a shorter one. What about $y_n=\log{x_n}$ then $$x_{n+2}=\sqrt{x_{n+1}x_n} \Rightarrow \\ \log{x_{n+2}}=\frac{\log{x_{n+1}}+\log{x_n}}{2} \Rightarrow \\ 2y_{n+2}=y_{n+1}+y_{n}$$ which is a linear homegenious recurrence which can be solved using characteristic polynomials. I.e. characteristic polynomial is $$2x^2-x-1=0$$ with $1$ and $-\frac{1}{2}$ as the roots. Thus the general term of the sequence is $$y_n=A\cdot 1^n+B\cdot \left(-\frac{1}{2}\right)^n= A+B\cdot \left(-\frac{1}{2}\right)^n$$ or $$x_n=e^{A}\cdot e^{B\cdot \left(-\frac{1}{2}\right)^n}\to e^A, n\to\infty$$ $A$ can be found from the initial $x_0,x_1$. A few examples here, here and here.

Answer 2

$x_{n+2} = \sqrt{x_{n+1} x_n} $.

Suppose $x_n =x_0^{a(n)}x_1^{b(n)} $ with $a(0) = 1, b(0) = 0, a(1) = 0, b(1) = 1 $. Then $x_0^{a(n+2)}x_1^{b(n+2)} =\sqrt{x_0^{a(n+1)}x_1^{b(n+1)}x_0^{a(n)}x_1^{b(n)}} =x_0^{(a(n+1)+a(n))/2}x_1^{(b(n+1)+b(n))/2} $ so that $a(n+2) =(a(n+1)+a(n))/2, b(n+2) =(b(n+1)+b(n))/2 $.

Both $a(n)$ and $b(n)$ are of the form $ru^n+sv^n$ where $u$ and $v$ are the roots of $x^2=(x+1)/2 $ or $2x^2-x-1=0 $, These are $x_{\pm} =\dfrac{1\pm\sqrt{1+8}}{4} =\dfrac{1\pm 3}{4} =1, -\dfrac12 $. We will use $u = 1, v = -\frac12$.

If $a(n) =r_au^n+s_av^n =r_a+s_a(-1/2)^n $ and $b(n) =r_bu^n+s_bv^n =r_b+s_b(-1/2)^n $ then, setting $n=0, 1$,

$1 = r_a+s_a,\\ 0 = r_a-s_a/2,\\ 0 = r_b+s_b,\\ 1 = r_b-s_b/2,\\ $

so $r_a = \frac13, s_a = \frac23, r_b = \frac23, s_b = -\frac23 $.

$\begin{array}\\ x_n &=x_0^{a(n)}x_1^{b(n)}\\ &=x_0^{r_a+s_a(-1/2)^n}x_1^{r_b+s_b(-1/2)^n}\\ &=x_0^{\frac13+\frac23(-1/2)^n}x_1^{\frac23-\frac23(-1/2)^n}\\ &\to x_0^{\frac13}x_1^{\frac23} \qquad\text{as } n \to \infty\\ \end{array} $

Answer 3

Here is as elementary a proof as I can come up with that the limit exists and the limit is $\sqrt[3]{x_1^2 x_0} $.

$x_{n+2} = \sqrt{x_{n+1} x_n} $, so, taking logs,

$\begin{array}\\ \log x_{n+2} &= \log\sqrt{x_{n+1} x_n}\\ &= \frac12 \log(x_{n+1} x_n)\\ &= \frac12 (\log x_{n+1} +\log x_n)\\ &= \frac12 \log x_{n+1} +\frac12 \log x_n\\ \end{array} $

Letting $y_n = \log x_n$, this becomes $y_{n+2} =\frac12 (y_{n+1}+y_n) $.

$y_{n+2}-y_{n+1} =\frac12 (y_{n+1}+y_n)-y_{n+1} =-\frac12(y_{n+1}-y_n) $.

By induction, for $k > 0$,

$\begin{array}\\ y_{n+2}-y_{n+1} &=-\frac12(y_{n+1}-y_n)\\ &=\frac14(y_{n}-y_{n-1})\\ &=-\frac18(y_{n-1}-y_{n-2})\\ &...\\ &=(-1)^k\dfrac1{2^k}(y_{n+2-k}-y_{n+1-k})\\ &=(-1)^{n+1}\dfrac1{2^{n+1}}(y_{1}-y_{0}) \qquad\text{by setting }k = n+1\\ &\to 0 \qquad\text{as } n \to \infty\\ \end{array} $

This shows that $\lim_{n \to \infty} y_n$ exists.

(Actually, a bit more is needed, but the following gives an explicit value so you don't need to worry about it.)

To get the value, sum the expression for $y_{n+1}-y_n$; all the intermediate terms will cancel out.

Write it in the form $y_{n}-y_{n-1} =(-1)^{n-1}\dfrac1{2^{n-1}}(y_{1}-y_{0}) =(-1)^{n-1}\dfrac{d}{2^{n-1}} $ where $d = y_{1}-y_{0} $.

$\begin{array}\\ y_m-y_1 &=\sum_{n=2}^m (y_n-y_{n-1})\\ &=\sum_{n=2}^m (-1)^{n-1}\dfrac{d}{2^{n-1}}\\ &=d\sum_{n=2}^m (-\frac12)^{n-1}\\ &=d\sum_{n=1}^{m-1} (-\frac12)^{n}\\ &=d\dfrac{-\frac12-(-\frac12)^m}{1-(-\frac12)}\\ &=d\dfrac{-\frac12-(-\frac12)^m}{\frac32}\\ &=d\dfrac{-1-2(-\frac12)^m}{3}\\ &=d(-\dfrac13-\dfrac23(-\frac12)^m)\\ &=-\dfrac{y_1-y_0}{3}-\dfrac{2(y_1-y_0)}{3}(-\frac12)^m\\ \end{array} $

so $y_m =\dfrac{2y_1+y_0}{3}-\dfrac{2(y_1-y_0)}{3}(-\frac12)^m \to\dfrac{2y_1+y_0}{3} $.

Therefore

$\begin{array}\\ \log x_m &\to \dfrac{2y_1+y_0}{3}\\ &= \dfrac{2\log(x_1)+\log(x_0)}{3}\\ &= \dfrac{\log(x_1^2 x_0)}{3}\\ &= \log(\sqrt[3]{x_1^2 x_0})\\ \end{array} $

so $x_m \to \sqrt[3]{x_1^2 x_0} $.

Answer 4

you dont need Lemma 4 to show the convergence. Lemma I-III show, that the even and uneven subsequence are bounded and monoton. Thus convergent. Now show that those limits are equal and you are done.(use the computational rules of limes) BTW: Im working on the same book at the moment, but I'm a chapter ahead of you :D