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


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!

  • $\begingroup$ maybe this problem have other methods. $\endgroup$ – math110 Nov 20 '13 at 12:55
  • $\begingroup$ @Ju'x that seems to prove that the sequence is bounded. I think more steps are needed to show that it converges. Correct? $\endgroup$ – apt1002 Nov 20 '13 at 13:05
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    $\begingroup$ @AlexR Let $a_1 = 0.25$ and $a_2 = 0.25$. Then $a_3 = 1$, and the sequence begins: $0.25, 0.25, 1, \ldots$ so that you cannot simply say it's "non-increasing." Indeed, it has increased here. The key observation is to think about when consecutive terms are less than $4$ vs. when they are greater than $4$; this will eventually happen barring a case where one of the terms is $4$. Then the sequence can be more carefully analyzed... $\endgroup$ – Benjamin Dickman Nov 20 '13 at 13:12
  • $\begingroup$ @apt1002: if $\limsup u_n \leq \ell \leq \liminf u_n$ then $u_n$ converges to $\ell$. $\endgroup$ – Siméon Nov 20 '13 at 13:20
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    $\begingroup$ @AlexR As commented previously, the analysis should be carried out for consecutive $a_k, a_{k+1}$ less than (or greater than) $4$. If you only write that $a_k < 4$, then perhaps we have a sequence beginning: $100, 1, 11, 1 + \sqrt{11}, \ldots$ where, strictly speaking, the sequence is neither non-decreasing nor non-increasing; we can see it has gone down, up, and down again at the start. The salient point is what happens to consecutive points eventually... $\endgroup$ – Benjamin Dickman Nov 20 '13 at 13:21

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 $$0<x_{n}<x_{n}+\dfrac{\sqrt{13}-1}{6}x_{n-1}=x_{n}-\dfrac{1-\sqrt{13}}{6}x_{n-1}\to 0$$


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 $x$.

  • $\begingroup$ Nice answer. +1. $\endgroup$ – Did Nov 20 '13 at 13:25
  • $\begingroup$ maybe this answer can use Monotone convergence theorem $\endgroup$ – math110 Nov 20 '13 at 13:55
  • $\begingroup$ @math110: why the need? $\endgroup$ – Siméon Nov 20 '13 at 14:01
  • $\begingroup$ @Siméon,I have post my solution $\endgroup$ – math110 Nov 29 '13 at 10:15

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 $$ \lim_{n\to\infty}a_n=4\tag{4} $$

  • $\begingroup$ Why $a_n > \sqrt{a_{n-1}}$? $\endgroup$ – Siméon Nov 29 '13 at 15:13
  • $\begingroup$ @Siméon: since $a_n=\sqrt{a_{n-1}}+\sqrt{a_{n-2}}$ and $a_{n-2}\gt0$. $\endgroup$ – robjohn Nov 29 '13 at 15:15
  • $\begingroup$ Oh right! Now I feel stupid. $\endgroup$ – Siméon Nov 29 '13 at 16:24

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.

  • $\begingroup$ This solution have some problem?@ Siméon $\endgroup$ – math110 Nov 29 '13 at 11:01

we have $$\sqrt{\min\left(\dfrac{a_{n+1}}{4},\dfrac{a_{n}}{4}\right)}\le\dfrac{a_{n+2}}{4}=\dfrac{\sqrt{\dfrac{a_{n+1}}{4}}+\sqrt{\dfrac{a_{n}}{4}}}{2}\le\sqrt{\max\left(\dfrac{a_{n+1}}{4},\dfrac{a_{n}}{4}\right)}$$ so $$\left(\min\left(\dfrac{a_{0}}{4},\dfrac{a_{1}}{4},\dfrac{a_{2}}{4}\right)\right)^{\frac{1}{2^{n+1}}}\le\dfrac{a_{n+2}}{4}\le\left(\max\left(\dfrac{a_{0}}{4},\dfrac{a_{1}}{4},\dfrac{a_{2}}{4}\right)\right)^{\frac{1}{2^{[\dfrac{n+1}{2}]}}}$$ so $$\lim_{n\to\infty}\dfrac{a_{n+2}}{4}=1$$ so $$\lim_{n\to\infty}a_{n+2}=4$$

I hope this methods is no problem.

  • $\begingroup$ Also, it is not clear what is the RHS of your second line and how you derive it from the first line. Actually, I don't think it is correct since we can find sequences $u_n$ that satisfy $\sqrt{\min(u_n,u_{n+1})}\leq u_{n+2} \leq \sqrt{\max(u_n,u_{n+1})}$ but do not converge. Example: $u_{2k}=0$ and $u_{2k+1}=1$. $\endgroup$ – Siméon Nov 29 '13 at 10:20

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