Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \ldots$ 
Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \dotsb$

Trial: Let
$$\begin{align}
P &= 3 \cdot 9^{\frac{1}{2}} \cdot 27^{\frac{1}{4}} \cdot 81^{\frac{1}{8}} \dotsb\\
\implies \ln P &=\ln3+\frac{1}{2} \ln 3^2+\frac{1}{2^2} \ln 3^3+\frac{1}{2^3} \ln 3^4 + \dotsb\\
&=\ln 3 \sum_{x=0}^{\infty}\frac{x+1}{2^x}
\end{align}$$
Then how I proceed. Is there any other simpler way to solve. Please help.

We have:
$$ P = \prod_{k=1}^\infty \big(3^k\big)^{\left(2^{1-k}\right)}$$
Take the $\log_3$ of both sides:
$$ \log_3 P = \sum_{k=1}^\infty \log_3 3^{k2^{1-k}}  $$
$$ \log_3 P = \sum_{k=1}^\infty k2^{1-k} $$
$$ \log_3 P = 2\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k$$
It is known that:
$$ \sum_{k=1}^{\infty} k r^k = \frac{r}{\left(1-r\right)^2} $$
So:
$$ \log_3 P = 2 \cdot \frac{\frac{1}{2}}{(1-1/2)^2} $$
$$ \log_3 P = \frac{1}{1/4} $$
$$ \log_3 P = 4 $$
$$ P = 3^4 = 81 $$
