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!


For every $x\in \mathbb R$ which $|x|\lt 1$, we have: $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ From here: $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ Now, multiplying $x^2$ in both side we get that: $$\sum_{n=1}^\infty nx^{n+1}=\frac{x^2}{(1-x)^2}$$ And so: $$\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n+1}=1$$
