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.

$$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$
therefore
$$\lim_{n\to \infty }\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots\times\sqrt[2^n]{2^n}=2^2=4$$
