Limit of $a_1=\sqrt{3}$, $a_2=\sqrt{3\sqrt{3}}$, $a_3=\sqrt{3\sqrt{3\sqrt{3}}}$... Find the limit of:
$a_1=\sqrt{3}$
$a_2=\sqrt{3\sqrt{3}}$
$a_3=\sqrt{3\sqrt{3\sqrt{3}}}$
...
By using induction, I found that the limit is 3 but it seems pretty strange to me, I thought it would go to a further number.
I just want to know if I'm right (proof-verification).

1)$3 > a_{i+1} > a_i > 1$.
Pf: by induction. 
Base step: $a_1 = \sqrt {3} > 1; a_1 = \sqrt{3} < 3$.
Induction: If $3> a_i > 1$ then $a_{i+1} = \sqrt{3*a_i} = \sqrt 3\sqrt{a_i}$
So $3 = \sqrt{3}\sqrt{3} > \sqrt{3}\sqrt{a_i} =a_{i+1}=\sqrt{3}\sqrt{a_i} > \sqrt{3}*1 > 1$.
And $a_{i+1} = \sqrt{3}\sqrt{a_i} > \sqrt{a_i}\sqrt{a_i} = a_i$.
So $3 > a_{i+1} > a+i > 1$.
2) So $a_i$ is increasing and bounded so $\lim\limits_{n\to\infty} a_n$ exists and is less than or equal to $3$.
I honestly have no idea whatsoever why you would have possibly thought the limit would be more than $3$.  
Let $\lim\limits_{n\to\infty} a_n = A$
Then $\lim\limits_{n\to \infty} \sqrt{3a_n} = \sqrt{3A}$
$\lim\limits_{n\to\infty}a_{n+1} = \sqrt{3A}$
$\lim\limits_{n+1\to\infty}a_{n+1} = \sqrt{3A}$
$A = \sqrt{3A}$
$A^2 = 3A$
$A =3$ or $A = 0$ but as $A > 1$ we have $A=3$.
