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).
 A: You can write $a_{n+1} = \sqrt{3a_n}$, so if we mark $ \displaystyle\lim _{n\to \infty}a_n$ we have   $$\lim _{n\to \infty}a_{n+1} = \lim _{n\to \infty}\sqrt{3a_n}$$
so $a^2 =3a$ so $a= 0$ or $a=3$. Since it is obviously increasing we have $a=3$.

Add to comment down.
It is obviously that $a_n>0$ for each $n$. Now prove that $a_n<3$ for each $n$ with induction. We do only induction step. So we know $a_n< 3$.
Then $a_{n+1}< \sqrt{3\cdot 3} =3$ and we are done. Now prove for increasing 
$$a_{n+1}-a_n = \sqrt{3a_n}-a_n = \sqrt{a_n}(\sqrt{3}-\sqrt{a_n}) >0 $$  
So sequence is increasing and bounded so it is convergent.
A: Suggestion:
$$\sqrt{3\sqrt{3\sqrt{3\sqrt{3\cdots}}}} =  3^{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots}$$
A: 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
A: 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 $A=3$.
