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

Now, one way to find this infinite sum is to convert it into an infinite sum of infinite geometric series:

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

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:
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.


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