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).

The sequence is defined recursively as
$x_0=\sqrt 3;\;x_n=\sqrt{3x_{n-1}}$
It's not difficult to prove that $x_n=3^{1-\frac{1}{2^{n+1}}}$ 
Indeed if $x=1$ we have $x_1=\sqrt{3\sqrt{3}}=\sqrt[4]{3^3}=3^{\frac34}=3^{1-\frac14}$
If the formula is valid for $n$ then $x_n=3^{1-\frac{1}{2^{n+1}}}$ we show that it is valid for $x_{n+1}=\sqrt{3x_n}=\sqrt{3\cdot 3^{1-\frac{1}{2^{n+1}}}}=\sqrt{3^{2-\frac{1}{2^{n+1}}}}=\left(3^{2-\frac{1}{2^{n+1}}}\right)^{\frac12}=3^{1-\frac{1}{2^{n+2}}}$
Therefore we proved by induction that $x^n=3^{1-\frac{1}{2^{n+1}}}$ and as $n\to\infty$ we have $x\to 3$
More quickly,assuming that the sequence converges,
$$x=\sqrt{3\sqrt{3\sqrt{3}}}\ldots$$
$$x^2=3\sqrt{3\sqrt{3}}\ldots$$
$$x^2=3x$$
$$x=3$$
hope it helps
