What is $2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty$ equal to?

Question

I came across this question while doing my homework:

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

$$\small\text{OR}$$

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

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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...

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How do you do it? Thanks!

Answer 1

Olympiad Tricks \begin{align*} y &= 1^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times \ldots \times 2^{\frac{n-1}{2^{n}}} \times \ldots \\[3pt] y^{2} &= 2^{\frac{1}{2}} \times 4^{\frac{1}{4}} \times 8^{\frac{1}{8}} \times \ldots \times 2^{\frac{n}{2^{n}}} \times \ldots \\[3pt] \frac{y^{2}}{y} &= \frac{2^{\frac{1}{2}} \times 4^{\frac{1}{4}} \times 8^{\frac{1}{8}} \times \ldots \times 2^{\frac{n}{2^{n}}} \times \ldots} {1^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 4^{\frac{1}{8}} \times \ldots \times 2^{\frac{n-1}{2^{n}}} \times \ldots} \\[3pt] y &= 2^{\frac{1}{2}} \times 2^{\frac{1}{4}} \times 2^{\frac{1}{8}} \times \ldots \times 2^{\frac{1}{2^{n}}} \times \ldots \\[3pt] &=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}+\ldots} \\[3pt] &=2^{1} \\[3pt] &=2 \end{align*}

Answer 2

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.

Answer 3

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: $$\sum_{n=1}^\infty n\left(\frac{1}{2}\right)^{n+1}=1$$