G.P.'s Power is G.P. itself

Slightly different from Nihar Mahajan 's solution.

$\begin{aligned} \text{Let } \Large x & \Large = \sqrt{3}\sqrt[4]{9} \sqrt[8]{27} \sqrt[16]{81}... = 3^{\frac{1}{2}} 3^{\frac{2}{4}} 3^{\frac{3}{8}} 3^{\frac{4}{16}} ... \\ & \Large = 3^{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16}+...} = 3^{ \sum_{k=1}^\infty {\frac{k}{2^k}}} \end{aligned}$

We note that:

$\begin{aligned} S & = \sum_{\color{#3D99F6}{k=1}}^\infty {\frac{k}{2^k}} = \sum_{\color{#3D99F6}{k=0}}^\infty {\frac{k}{2^k}} \\ S & = \sum_{\color{#3D99F6}{k=1}}^\infty {\frac{k}{2^k}} = \sum_{\color{#3D99F6}{k=0}}^\infty {\frac{\color{#3D99F6}{k+1}}{2^{\color{#3D99F6}{k+1}}}} \end{aligned}$

Now, we have:

$\begin{aligned} S & = 2S - S \\ & = 2 \sum_{k=0}^\infty {\frac{k+1}{2^{k+1}}} - \sum_{k=0}^\infty {\frac{k}{2^k}} \\ & = \sum_{k=0}^\infty {\frac{k+1}{2^{k}}} - \sum_{k=0}^\infty {\frac{k}{2^k}} \\ & = \sum_{k=0}^\infty {\frac{1}{2^{k}}} = \frac{1}{1-\frac{1}{2}} = 2 \end{aligned}$

$\Rightarrow x = 3^S = 3^2 = \boxed{9}$

Let us write the product like this:

$3^{\left(\dfrac{1}{2}\right)}.3^{\left(\dfrac{2}{4}\right)}.3^{\left(\dfrac{3}{8}\right)}.3^{\left(\dfrac{4}{16}\right)}... \\ = 3^{\left(\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\dfrac{4}{16}+...\right)} \\ = 3^{\left(1\times\dfrac{1}{2}+2\times\dfrac{1}{4}+3\times\dfrac{1}{8}+4\times\dfrac{1}{16}+...\right)}$

So we need to evaluate the following infinite sum:

$1\times\dfrac{1}{2}+2\times\dfrac{1}{4}+3\times\dfrac{1}{8}+4\times\dfrac{1}{16}...$

We find that we have an Arithmetic Progression with first term $(a=1)$ and difference $(d=1)$ .

We also have a Geometric progression with first term $\left(b=\dfrac{1}{2}\right)$ and ratio $\left(r=\dfrac{1}{2}\right)$

We observe that this is an Arithmetic-o-geometric progression whose infinite sum is given by :

$\dfrac{ab}{1-r}+\dfrac{dbr}{(1-r)^2}$

Substituting the given values we have:

$\dfrac{(1)\left(\dfrac{1}{2}\right)}{\left(1-\dfrac{1}{2}\right)}+\dfrac{(1)\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)}{\left(1-\dfrac{1}{2}\right)^2} \\ = \dfrac{\left(\dfrac{1}{2}\right)}{\left(\dfrac{1}{2}\right)} +\dfrac{\left(\dfrac{1}{2}\right)^2}{\left(\dfrac{1}{2}\right)^2} \\ = 1+1 \\ =\boxed{2}$

So the given sum $=3^2 = \boxed{9}$