Find $\lim_{n \to \infty } a_n$ \begin{cases} a_1=\sqrt 3 \\ a_2 = \sqrt {3\sqrt 3}\\ a_n = \sqrt {3a_{n-1}} \quad \text{for } n\in\mathbb Z^+\end{cases}
This sequence is bounded above by $3$ and is monotone increasing, so by monotone bounded sequence theorem, the sequence converges.  
But, the question asks to find $\lim_\limits{n \to \infty } a_n$. I guess the limit is $3$, but don't know how to prove it. 
Could you give some hint? Thank you in advance. 
 A: $$a_n=3^{1/2+1/4+\cdots+1/2^n}$$
Now $S(n)=\dfrac12+\dfrac14+\cdots+\dfrac1{2^n}=\dfrac12\left(\dfrac{1-\left(\dfrac12\right)^n}{1-\dfrac12}\right)$
$\lim_{n\to\infty}S(n)=\dfrac12\left(\dfrac{1-0}{1-\dfrac12}\right)=1$
A: Hint 
If $\ell$ is the limit, then $$\ell=\sqrt{3\ell}.$$
A: Suppose $a_n\to l$ for $n\to \infty$. Then also $a_{n+1}\to l$ because $n+1\sim n$ as $n\to\infty$. So
$$\lim a_n=\lim a_{n+1}=\lim3\sqrt{a_n}\implies l=3\sqrt{l}$$
A: Suppose that $a$ is the limit. Then:
$$a = \sqrt{3\sqrt{3\sqrt{3\sqrt{\ldots}}}}.$$
You can write that:
$$a = \sqrt{3a}.$$
This means that:
$$a = \sqrt{3}\sqrt{a} \Rightarrow \sqrt{a} = \sqrt{3} \Rightarrow a = 3.$$
Anyway, we don't know if this sequence will converge to the limit $a$.
To this aim, notice that:
$$\frac{\partial a_{n}}{\partial a_{n-1}} = \frac{1}{2}\sqrt{\frac{3}{a_{n-1}}}.$$
For $a_{n-1} = a = 3$, we get that the value of this derivative is $\frac{1}{2}$ which is in modulus less than $1$. Then, the sequence converge to $a= 3$.
A: Set $b_n := \ln a_n$:


*

*$b_1 = \frac{1}{2}\ln 3$

*$b_{n} = \frac{1}{2}\ln{(3a_{n-1})} = \frac{1}{2}\ln 3 + \frac{1}{2}\ln a_{n-1} = \frac{1}{2}\ln 3 + \frac{1}{2}b_{n-1}$

*$\Rightarrow b_n = \left(\frac{1}{2}+\cdots + \frac{1}{2^n} \right)\ln 3= \left(1- \frac{1}{2^n} \right)\ln 3 \stackrel{n \rightarrow \infty}{\rightarrow}\ln 3$

*$a_n = e^{b_n} \stackrel{n \rightarrow \infty}{\rightarrow} e^{\ln 3} = 3$

