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I believe this is a very simple one, but I simply can't figure it out. How to solve?

$$\frac12\cdot\frac34\cdot\frac56\cdots\frac{17}{18}\cdot\frac{19}{20}$$

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  • $\begingroup$ You can directly calculate it. It is not that extreme at all. $\endgroup$
    – SarpSTA
    Jan 13, 2016 at 8:39
  • $\begingroup$ We can think of it as $\binom{20}{10}\frac{1}{2^{20}}$, the probability of $10$ heads in $20$ tosses of a fair coin. $\endgroup$ Jan 13, 2016 at 8:48

3 Answers 3

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The general compact writing of this sort of fraction is $$\frac{1\cdot 3\cdot5\dotsm(2n-1)}{2\cdot 4\cdot6\dotsm2n}=\frac{(2n)!}{(2\cdot 4\cdot6\dotsm2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}=\frac1{2^{2n}}\binom{2n}{n}.$$

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  • $\begingroup$ nice...............+1 $\endgroup$ Jan 13, 2016 at 9:32
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Check out the Wallis product https://en.wikipedia.org/wiki/Wallis_product and think square root.

The product of even numbers is a factorial of half the largest times a power of 2, for the odd numbers you may add even factors (that you just learned how to write in a compact way) to get a factorial.

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Bernard gave the compact formula $$A_n=\frac{1\cdot 3\cdot5\dotsm(2n-1)}{2\cdot 4\cdot6\dotsm2n}=\frac{(2n)!}{2^{2n}(n!)^2}$$ Now, let me assume that you have large values of $n$ and that you need a shortcut approximation of the result in a decimal form.

The factorial can be evaluated using Stirling approximation $$p!\approx \sqrt{2\, \pi\, p}\, \Big(\frac p e\Big)^n$$ Applying it to $A_n$ gives $$A_n\approx \frac{1}{\sqrt{\pi n} }$$

Applied to $n=10$ as in your post, this will give a value of $0.178412$ while the true value would be $0.176197$ which is not too bad (I hope). If you want something better, take an extra term in Stirling formula to get $$A_n\approx \frac{1}{\sqrt{\pi n} }\big(1-\frac 1 {8n}\big)$$ which would give $0.176182$.

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