Krdytkyu

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

Hint:

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

Now $displaystylesum_{r=1}^nleft(dfrac13right)r=dfrac13left(dfrac{1-left(dfrac13right)^n}{1-dfrac13}right)$^

Finally, $displaystylelim_{ntoinfty}left(dfrac13right)^n=?$

Use geometric summation:

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

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