\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.
Answer
Hint
If $\ell$ is the limit, then $$\ell=\sqrt{3\ell}.$$
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