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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 $\lim_{n \to \infty}{3^{(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + ... + \frac{1}{3^n})}} = \lim_{n \to \infty}{3^\frac{1}{2}} = \sqrt{3}$

envy grunt
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    $3^{1/3+...+1/3^{3^n}}$ – dmtri Nov 17 '18 at 09:56
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    I do not think it is the best approach, but if $L=\lim\limits_{n \to \infty}\left(\sqrt[3]{3 \sqrt[3]{3 \sqrt[3]{3...}}}\right)$ exists then $L=\sqrt[3]{3L}$ so $L^3 =3L$ and so $L=0$ or $\pm\sqrt{3}$ or perhaps $\pm\infty$, giving $L=\sqrt{3}$ as the only positive finite solution – Henry Nov 17 '18 at 10:09
  • yes, it's more complicated than submitted solution, but also works fine – envy grunt Nov 17 '18 at 10:12

3 Answers3

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Hint:

$$\prod_{r=1}^n3^{(1/3)^r}=3^{\sum_{r=1}^n(1/3)^r}$$

Now $\displaystyle\sum_{r=1}^n\left(\dfrac13\right)r=\dfrac13\left(\dfrac{1-\left(\dfrac13\right)^n}{1-\dfrac13}\right)$^

Finally, $\displaystyle\lim_{n\to\infty}\left(\dfrac13\right)^n=?$

lab bhattacharjee
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Use geometric summation:

$$\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})}=\\e^{\log(\lim_{n \to \infty}{(\sqrt[3]{3} * \sqrt[9]{3} * ... * \sqrt[3^n]{3})})}=\\e^{\lim_{n \to \infty}\sum _{i=1}^n 3^{-i} \log (3)}=\\e^{\sum _{i=1}^\infty 3^{-i} \log (3)}=\\e^{\frac{\log (3)}{2}}=\sqrt{3}$$

MeMyselfI
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$$\sqrt[3]{3 * \sqrt[3]{3 * \sqrt[3]{3 * ...}}} = n$$ $$\sqrt[3]{3 * n} = n$$ $$3*n=n^3$$ $$3=n^2$$ $$n=\sqrt{3}$$

Cahid Enes Keles
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