find the limits of the given sequence?

Question

the recursive sequencen ${a_n}$ is given by setiing $ a_1= 1,a_2 =2$ and $a_{n+1} =(a_{n-1})^{1/2} +({a_n})^{1/2} $ for $n\ge2$.

$a)$show the sequence is $bounded$ and strictly increasing .

$b)$ find its limits ?

my answer : for option $a)$ ..we have $a_1 <a_2 <a_3 $and for any $ n\in N$...... if $a_n< a_{n+1} <a_{n+2} $,then $a_{n+2} < a_{n+3}$ now by mathematical induction sequence the sequnce ${a_n}$ is strictly increasing...

here im struck at option $b)$,,,,

Any hints or solution can be aappreciated,,as

pliz help me..

Answer 1

The limit is $4$, because, if $l$ is the limit, then\begin{align}l&=\lim_{n\in\mathbb N}a_n\\&=\lim_{n\in\mathbb N}\sqrt{a_{n-1}}+\sqrt{a_{n-2}}\\&=2\sqrt l.\end{align}Therefore, $l=0$ or $l=4$. But it can't be $0$ (each term is equal or greater than $1$).

I don't now if you proved that the sequence is bounded, but that's not hard. The first two terms are smaller than $4$. And if $a_{n-1},a_{n-2}<4$, then$$a_n=\sqrt{a_{n-1}}+\sqrt{a_{n-2}}<\sqrt4+\sqrt4=4.$$

Answer 2

You have already proved that the sequence is monotone.

The boundedness is simple, so there is limit which we call it $a$

In order to find the limit we solve $$ lim _{n\to \infty} a_{n+1} = lim _{n\to \infty} (a_{n-1})^{1/2} + lim _{n\to \infty} ({a_n})^{1/2}$$

$$ a=2\sqrt a \implies a^2=4a$$

Since a>0, we get $$a=4$$