Evaluate $\sqrt[2]{2} \cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots$ Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$
My attempt:First solve when $n$ is not infinity then put infinity in.
$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$
$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$
Now calculate the sum of the powers:
$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$
$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$
Now calculate the numerator:
$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$
$$+$$
$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$
$$+$$
$$\vdots$$
$$+$$
$$2^0=2^1-1$$
$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$
Now put the numerator on the fraction:
$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$
Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$
Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.
That gives the total answer is $4$.
But now they are two problems:

1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.
2.My answer is too long.

Now I want you to help me with these problems.Thanks.

$$\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots$$
When we use the LOG function get (when $n$ is positive):
$$\ln\left(\left(2^n\right)^{\frac{1}{2^n}}\right)=\frac{n\ln(2)}{2^n}$$
And using, when $a$ and $b$ are positive:
$$\ln(ab)=\ln(a)+\ln(b)$$
So, we get that:
$$\ln(\text{P})=\frac{1\ln(2)}{2^1}+\frac{2\ln(2)}{2^2}+\frac{3\ln(2)}{2^3}+\dots=\ln(2)\left[\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots\right]$$
And, now we know that:
$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots=\frac{2}{(2-1)^2}=2$$
Using, when $|x|>1$:
$$\sum_{n=1}^{\infty}\frac{n}{x^n}=\frac{x}{(x-1)^2}$$
So:
$$\ln(\text{P})=\ln(2)\left[2\right]=2\ln(2)\Longleftrightarrow\text{P}=4$$
We, find the answer:
$$\color{red}{\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots=4}$$
