What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to? I came across this question while doing my homework: 

$$\Large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=?$$

$$\small\text{OR}$$
$$\large\prod\limits_{x=1}^{\infty} (2x)^{\frac{1}{4x}} = ?$$

My Attempt:
$\large 2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$
$\large \Rightarrow 2^{\frac{1}{4}}\cdot2^{\frac{2}{8}}\cdot2^{\frac{3}{16}}\cdot2^{\frac{4}{32}}\cdots\infty$
$\large \Rightarrow 2^{(\frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \infty)}$
OR, $\large 2^{\space (\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}})}$
$\cdots$ That's it ... I am stuck here ... It does not resemble any series I know...


How do you do it? Thanks!


The only thing left that we have to do is to evaluate the following infinite sum:
$$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+1}}$$
Dividing by $2$ would gives us $$\sum\limits_{x=1}^{\infty} \frac{x}{2^{x+2}}=\sum\limits_{x=2}^{\infty} \frac{x-1}{2^{x+1}}$$
Now, subtract this from the original equation. The limit should now not be too hard to find. 
