Easy nested sequence

Let $\sqrt [3]{3 \sqrt [3]{3 ..\ldots}} = x$ . Then $x^3 = 3 \sqrt [3]{3 \sqrt [3]{3 ..\ldots}}$ .

To simplify, we get $x^3 = 3x$

Upon further simplification, we get:

$x^2 = 3$ or $x = \sqrt {3}$ .

Note that we couldn't possibly have a negative answer since we are working with only positive numbers. Also, in $x^3 = 3x$ can not result in $x = 0$ as $0^3 \neq 3$ .

First, you cube both sides

${ \left[ \sqrt [ 3 ]{ 3\cdot \sqrt [ 3 ]{ 3\cdot \sqrt [ 3 ]{ 3\cdot } .. } } \right] }^{ 3 }={ \left( \sqrt { n } \right) }^{ 3 }$

We know that $\sqrt [ 3 ]{ 3\cdot \sqrt [ 3 ]{ 3\cdot \sqrt [ 3 ]{ 3\cdot } .. } } =\sqrt { n }$

You substitute that to your equation and you will get

${ \left[ \sqrt [ 3 ]{ 3\cdot \sqrt { n } } \right] }^{ 3 }={ \left( \sqrt { n } \right) }^{ 3 }$

$3\cdot \sqrt { n } ={ n }^{ \frac { 3 }{ 2 } }$

$3=\frac { { n }^{ \frac { 3 }{ 2 } } }{ { n }^{ \frac { 1 }{ 2 } } } ={ n }^{ \frac { 2 }{ 2 } }=n$

Therefore
$n=3$

I DIDN'T WANT TO JUSTIFY WHY ZERO IS NOT A POSSIBLE SOLUTION, SO I USED LOGARITHM

Given: $\sqrt[3]{3\sqrt[3]{3\sqrt [3]{3\ldots}}}$ = $\sqrt n$
Taking Logarithm on both sides,
$\log \sqrt[3]{3\sqrt[3]{3\sqrt [3]{3\ldots}}}=\log n^{\frac{1}{2}}$
$\Rightarrow \frac{1}{3}\times \log 3\sqrt[3]{3\sqrt [3]{3\ldots}}=\frac{1}{2}\times \log n$
$\Rightarrow \log 3\sqrt[3]{3\sqrt [3]{3\ldots}}=\frac{3}{2}\times \log n$
$\Rightarrow \log 3 +\log \sqrt[3]{3\sqrt [3]{3\ldots}}=\frac{3}{2}\log n$
Replacing $\sqrt[3]{3\sqrt [3]{3\ldots}}$ with $n^\frac{1}{2}$ on L.H.S ,
$\Rightarrow \log 3 +\log n^\frac{1}{2}=\frac{3}{2}\log n$
$\Rightarrow \log 3 +\frac{1}{2}\log n=\frac{3}{2}\log n$
$\Rightarrow \log 3=\frac{3-1}{2}\log n$
$\Rightarrow \log 3=\frac{2}{2}\log n$
$\Rightarrow \log 3=\log n$
$\Rightarrow \boxed{3=n}$
So, $\boxed{n=3}$

3^(1/3+1/9+1/27......)=3^(1/3)/(1-1/3)...(using GP Series)=3^1/2

it is 3^(1/3+ 1/9.......................) solving g.p. expression we get 3^0.5 so n=3

We can write the question as $\color{#20A900}{\sqrt[3]{3\sqrt{n}}}$ because the question is embedded in itself.Raise both sides to the 6th power:

$\color{#D61F06}{(\sqrt[3]{3\sqrt{n}})^6=(\sqrt{n})^6\\9n=n^3\\=n^3-9n=n(n^2-9)=n(n+3)(n-3)=0}$ Cube root of a positive number is a positive number which eliminates 0 and -3 and leaves $\color{#3D99F6}{n=\boxed{3}}$

$\begin{aligned} \sqrt [ 3 ]{ 3\sqrt [ 3 ]{ 3\sqrt [ 3 ]{ 3\cdots } } } & = & \sqrt { n } \\ { (\sqrt [ 3 ]{ 3\sqrt { n } } ) }^{ 6 } & = & { (\sqrt { n } ) }^{ 6 } \\ 9n & = & { n }^{ 3 } \\ { n }^{ 3 }-9n & = & 0 \\ n(n-3)(n+3) & = & 0 \\ n & = & \boxed { 3 } \quad (n>0) \end{aligned}$

We have $\sqrt[3]{3 \sqrt[3]{3 \ldots}} = \sqrt{n}$ , so therefore, $\sqrt{n}= \sqrt[3]{{3\sqrt{n}}}$ . Then $n\sqrt{n} = 3\sqrt{n}$ , and from this, you know $n = 3$ .