Prove that the succession $x_{n}$ where $x_{1}=1$ and $x_{n+1} = \sqrt{3x_{n}}$ is convergent Well, ive been having a weee bit of problem solving this homework, can anyone give me a hand?
Prove that the sequence $x_{n}$ where $x_{1}=1$ and $x_{n+1} = \sqrt{3x_{n}}$ is convergent and calculate its limit
 A: Hint:
Prove the 3 is an upper bound (by induction:if $x_{n}<3$ then $x_{n+1}<3$)
Prove the whole sequence is strictly increasing (ie. $x_{n+1}>x_{n}$, use the fact that $3$ is an upper bound).
Apply monotone convergence to know that it does converge to some $L$.
Use limit arithmetic to show that $L$ behave the same way the sequence behave, that is $L=\sqrt{3L}$.
A: Hint $x_n=\sqrt{3 \sqrt{3 \ldots\sqrt{3}}}  =3^{\frac12+\frac14+\frac18+\cdots+\frac{1}{2^{n}}}$
A: Hints:
1) By induction you can prove that $1\leq x_n\leq 3$.
2) If it converges to $x$ then $x=\lim x_{n+1}=\lim\sqrt{3x_{n}}=\sqrt{3x}$.
A: To prove convergence, prove the sequence is a contraction. See here. To find the limit, solve the equation
$$ x=\sqrt{3x} .$$
A: Hints.
By induction (show the first case):
$$\begin{align*}(1)&\;\;x_n\le 3\;\;\forall\,n\in\Bbb N\;:\;x_n:=\sqrt{3x_{n-1}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{3\cdot3}=3\\{}\\(2)&\;\;x_n\le x_{n+_1}\;:\;x_n:=\sqrt{3x_{n-1}}\stackrel{\text{Ind. Hyp.}}\le\sqrt{3x_n}=:x_{n+1} \end{align*}$$
A: Let $y_n=\dfrac{x_n}3$, then 
$$y_{n+1}=\sqrt{y_n}$$
so
$$y_{n+1}=\sqrt{y_n}={(y_{n-1})}^{\frac14}=\dotsb=$$
