## Monday Math 161

Find the infinite product $2^{\frac12}\cdot4^{\frac14}\cdot8^{\frac18}\cdots=\prod\limits_{n=1}^{\infty}\left(2^n\right)^{\frac1{2^n}}$

First, we convert all of these to powers of two: $\left(2^n\right)^{\frac1{2^n}}=2^{\frac{n}{2^n}}$, so that our product is $2^{\frac12}\cdot2^{\frac24}\cdot2^{\frac38}\cdots$. And by the basic law of exponents $b^m\cdot{b^n}=b^{m+n}$, we thus have
$\prod\limits_{n=1}^{\infty}2^{\frac{n}{2^n}}=2^{\sum\limits_{n=1}^{\infty}\frac{n}{2^n}}$

Now, one way to find this infinite sum is to convert it into an infinite sum of infinite geometric series:
$\begin{array}{r}\frac12+\frac14+\frac18+\frac1{16}+\cdots\\\frac14+\frac18+\frac1{16}+\cdots\\\frac18+\frac1{16}+\cdots\\\frac1{16}+\cdots\\\vdots\\\hline\\\frac12+\frac24+\frac38+\frac4{16}+\cdots\end{array}$

And using the infinite geometric series formula $a_0+a_0r+a_0r^2+a_0r^3+\cdots=\frac{a_0}{1-r}$ with our $r=\frac12$, we see
$\begin{array}{rcl}\frac12+\frac14+\frac18+\frac1{16}+\cdots&=&1\\\frac14+\frac18+\frac1{16}+\cdots&=&\frac12\\\frac18+\frac1{16}+\cdots&=&\frac14\\\frac1{16}+\cdots&=&\frac18\\\vdots&=&\vdots\\\hline\\\frac12+\frac24+\frac38+\frac4{16}+\cdots&=&1+\frac12+\frac14+\frac18+\cdots\end{array}$

Which is itself a geometric series with sum $\frac{1}{1-\frac12}=2$
And thus our product is $\prod\limits_{n=1}^{\infty}2^{\frac{n}{2^n}}=2^{\sum\limits_{n=1}^{\infty}\frac{n}{2^n}}=2^2=4$

There is also another way we could have found our sum, using calculus. Noting that
$\sum\limits_{n=0}^{\infty}x^n=\frac{1}{1-x}$ for $\left|x\right|<1$, we see that
$\sum\limits_{n=1}^{\infty}\frac{x^n}{2^n}=\frac{\frac{x}2}{1-\frac{x}2}=\frac{x}{2-x}$ for $\left|x\right|<2$
Now, taking the derivative of both sides with respect to x:
$\sum\limits_{n=1}^{\infty}\frac{nx^{n-1}}{2^n}=\frac{2}{\left(2-x\right)^2}$
But the left hand sum becomes our desired sum when x=1, so making that substitution, we see $\sum\limits_{n=1}^{\infty}\frac{n}{2^n}=\frac{2}{\left(2-1\right)^2}=2$,
the same result we found above.