<s>[INST] Find a limit of $\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3}$ I have no idea how can i solve this. When i'm trying to transform a multiplication I always get a $0*\infty$ ambiguity.
I found only that
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}})$
EDIT: Solution
$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})} = \lim_{n \to \infty}(\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3*...}}}) = \lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}}$
$\sum_{r=1}^n{\frac{1}{3^n}} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{1}{2}$
So $\lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}} = \lim_{n \to \infty}{3^\frac{1}{2}} = \sqrt{3}$
 [/INST] $$\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3 * ...}}} = n$$
$$\sqrt[3]{3 * n} = n$$
$$3*n=n^3$$
$$3=n^2$$
$$n=\sqrt{3}$$
</s>