How to prove this limit exists: $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$

Question:

Consider a sequence $\{a_{n}\}$ such that $a_{1},a_{2}>0$, and for all $n \in \mathbb{N}$ we have:

$$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$$

Prove that :

$\displaystyle\lim_{n\to\infty}a_{n}$ exists and find this limit.

My work: If this limit exists, let $\displaystyle\lim_{n\to\infty}a_{n}=x>0$; then we have $$x=\sqrt{x}+\sqrt{x}\Longrightarrow x=4$$

But I can't see how to prove the limit of $\{a_{n}\}$ exists.

Thank you for you help!

Solution 1:

Obviously, it is enough to show that

$$ \limsup a_n \leq 4 \leq \liminf a_n. $$

To prove this, first remark that the sequence $M_n = \max\{4,a_n, a_{n-1}\}$ is non-increasing. Indeed, the trivial lower bound $M_n \geq 4$ yields $a_{n+1} \leq 2\sqrt{M_n} \leq M_n$ ; we conclude with $$ M_{n+1} = \max\{4, a_{n+1}, a_n\} \leq \max\{4, M_n, M_n\} = M_n. $$ As a consequence, an upper bound for $a_n$ is $\max\{4, a_1, a_2\}$. In the same way, we can prove the lower bound $a_n \geq \min\{4, a_1, a_2\}$.

These bounds show that both $\liminf a_n$ and $\limsup a_n$ are finite and positive. Furthermore we deduce from $a_{n+1} = \sqrt{a_n} + \sqrt{a_{n-1}}$ that $$ \liminf a_n \geq 2\sqrt{\liminf a_n},\qquad \limsup a_n \leq 2\sqrt{\limsup a_n} $$ The conclusion follows because $x \geq 2\sqrt{x} \implies x \geq 4$ as well as $x \leq 2\sqrt{x}\implies x \leq 4$ for every positive real number $x$.

Solution 2:

and I have consider other solution

if $a_{0}>0,a_{1}>0,$,and $a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}$,then $a_{n}\to 4$

pf: if $0<a_{0}\le a_{1}\le 1$,then $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{1}}\ge a_{1}$$ we note $$a_{n+2}-a_{n+1}=(\sqrt{a_{n+1}}+\sqrt{a_{n}})-(\sqrt{a_{n}}+\sqrt{a_{n-1}})=(\sqrt{a_{n+1}}-\sqrt{a_{n}})+(\sqrt{a_{n}}-\sqrt{a_{n-1}})\ge 0$$ so $\{a_{n}\}$is Monotone increasing,

$$a_{n}\le 4,\mbox{since } a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}\le 2+2=4$$ (2):if $0<a_{1}\le a_{0}\le 1$,then $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}\ge a_{0}$$ and $$a_{n+2}-a_{n+1}=\sqrt{a_{n+1}}-\sqrt{a_{n-1}}\ge 0$$.the simalar we have $\{a_{n}\}$ is increasing and $a_{n}\le 4$.

(3):if$a_{0}>1$(or $a_{1}>1$),then we have $$a_{2}=\sqrt{a_{1}}+\sqrt{a_{0}}>1,\Longrightarrow a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_{n}}>1$$ let $x_{n}=|a_{n}-4|$,then we have $$x_{n+1}=|a_{n+1}-4|<|\sqrt{a_{n}}-2|+|\sqrt{a_{n-1}}-2|=\dfrac{|a_{n}-4|}{\sqrt{a_{n}}+2}+\dfrac{|a_{n-1}-4|}{\sqrt{a_{n-1}}+2}<\dfrac{1}{3}x_{n}+\dfrac{1}{3}x_{n-1}$$ so $$x_{n+1}-\dfrac{1-\sqrt{13}}{6}x_{n}<\dfrac{1+\sqrt{13}}{6}(x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1})<\cdots<\left(\dfrac{1+\sqrt{13}}{6}\right)^n\left(x_{1}-\dfrac{1-\sqrt{13}}{6}x_{0}\right)\to 0,n\to\infty$$ so $$0<x_{n}<x_{n}+\dfrac{\sqrt{13}-1}{6}x_{n-1}=x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1}\to 0$$

Solution 3:

First note that since $a_n\gt0$, $$ \begin{align} a_n-4 &=(\sqrt{a_{n-1}}-2)+(\sqrt{a_{n-2}}-2)\\ &=\frac{a_{n-1}-4}{\sqrt{a_{n-1}}+2}+\frac{a_{n-2}-4}{\sqrt{a_{n-2}}+2}\tag{1} \end{align} $$ Furthermore, $$ a_n\gt\sqrt{a_{n-1}}\tag{2} $$ Therefore, since $a_0\gt0$, there is an $n_0$ so that for $n\ge n_0$, $a_n\gt\frac14$. Thus, for $n\ge n_0+2$, $(1)$ says $$ |a_n-4|\lt\tfrac25\big(|a_{n-1}-4|+|a_{n-2}-4|\big)\tag{3} $$ Since both roots of $x^2-\tfrac25x-\tfrac25=0$ have absolute value less than $1$, by comparison with any positive sequence so that $b_n=\tfrac25(b_{n-1}+b_{n-2})$, $(3)$ implies that $$ \lim_{n\to\infty}a_n=4\tag{4} $$

Solution 4:

other idea:

Now I have edit:

if $c_{0}>0,c_{1}>0.c_{n+1}=\sqrt{c_{n}}+\sqrt{c_{n-1}},n\ge 1$, then $\lim_{n\to\infty}c_{n}$ is exsit.

pf:let $$a_{0}=\min(c_{0},c_{1},4),b_{0}=\max(c_{0},c_{1},4)$$ then $$a_{n}=2\sqrt{a_{n-1}},b_{n}=2\sqrt{b_{n-1}}$$

then $$a_{0}\le a_{1}\le\cdots\le 4,b_{0}\ge b_{1}\ge \cdots\ge 4$$ and since $a_{0}\le c_{0},a_{0}\le c_{1}$,so $$a_{0}\le min(c_{0},c_{1})$$ Aussmu that for $n-1$,we have $$a_{n-1}\le\min(c_{2n-2},c_{2n-1})$$ then $$c_{2n}=\sqrt{c_{2n-1}}+\sqrt{c_{2n-2}}\ge 2\sqrt{a_{n-1}}=a_{n}$$ $$c_{2n+1}=\sqrt{c_{2n}}+\sqrt{c_{2n-1}}\ge\sqrt{a_{n}}+\sqrt{a_{n-1}}\ge 2\sqrt{a_{n-1}}=a_{n}$$

so According to the mathematical induction we have $a_{n}\le \min(c_{2n},c_{2n+1})$.and simaler we have $$b_{n}\ge \max(c_{2n},c_{2n+1})$$ so $$a_{n}\le c_{2n}\le b_{n},a_{n}\le c_{2n+1}\le b_{n}$$

By done.