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.
 A: I hope this would be useful for you.
So your trying to evaluate $\displaystyle\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}$. Consider first $\displaystyle\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}$, after developing the product properly. Now you have to make sense of $$\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}$$
, so to reach a specific value, note that $\sum_{n=1}^{k}\frac{n}{2^{n}}=\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}$, therefore 
\begin{eqnarray*}
\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}&=&\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}
{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}\\
&=&\lim_{k\to\infty}-\frac{k}{2^{k}}+\frac{1-\frac{1}{2^{k+1}}}{1-\frac{1}{2}}\\
&=&2
\end{eqnarray*}
Finally $$\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{2}=4$$
A: $$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$
$$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$
$$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$
$$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$
therefore
$$\lim_{n\to \infty }\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots\times\sqrt[2^n]{2^n}=2^2=4$$
A: $$\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots$$
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:
$$\color{red}{\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots=4}$$
A: I believe this is a "known" representation of the key summation:


$$\sum_{n=1}^{\infty} \frac{n}{2^n} \;=\; \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \cdots \;=\; 2 $$

A: $I=1+\frac 12+\frac 14+\frac 18+\frac 1{16}+\cdots=2$
$I^2 =1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$
$I^2 -I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots=2^2-2=2$
Then
$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots\cdot \sqrt[2^n]{2^n}=2^{I^2-I}=2^2=4$
A: A standard thing to do to, is too use the commonly known geometric series formula,  and set:
$$f(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$$
The formula is not hard to prove, and it holds for $|x|<1$ (which can be seen by the ratio test for convergence) we take $x \in (-1,1)$.
Then by term by term differentiation we get:
$$f'(x)=\sum_{n=1}^{\infty} nx^{n-1}=\frac{1}{x}\sum_{n=1}^{\infty} nx^n=\frac{1}{(1-x)^2}$$
Hence we have:
$$\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$
And the case your interested about is $x=\frac{1}{2}$
A: I would like to generalize Behrouz' rather nifty answer:
$$I=\sum_{n=1}^{\infty} \frac{n}{r^n}=\frac{1}{r}+\frac{2}{r^2}+\frac{3}{r^3}+\frac{4}{r^4}$$
The key is now to multiply everything by $r$:
$$rI=1+\frac{2}{r}+\frac{3}{r^2}+\frac{4}{r^3}$$
Now we calculate $rI-I$
$$rI-I=1+(\frac{2}{r}-\frac{1}{r})+(\frac{3}{r^2}-\frac{2}{r^2})+(\frac{4}{r^3}-\frac{3}{r^3})+...$$ simplifies to
$$1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}...$$ And that is our good old geometric sum 
