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$$\prod_{n=1}^\infty\left(\frac{3}{2^n}\right)^{1/2^n} $$

Sorry if that's not clear, that's $(1/2^n)$ in the superscript. The answer is $3/4$. I've never worked with infinite products before, so a thorough explanation would be much appreciated. Thank you.

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It might help to convert to an infinite sum using logarithms: $\prod_{n = 1}^\infty a_n = \operatorname{exp}(\sum_{n =1 }^\infty \ln a_n)$ – Austin Mohr Feb 13 '14 at 6:53

1 Answer 1

If you take the logarithm, you get: $$\sum 2^{-n}\log\left(3\cdot 2^{-n}\right)= \sum 2^{-n}\left(\log3-n\log 2\right)=\log 3 -2\log2$$ Using the fact that $\sum n2^{-n}=2$ and $\sum 2^{-n}=1$.

Now exponentiate to get $3/4$.

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