$$\lim_{n\rightarrow\infty}\left( \frac{1*3*5*...*(2n-1) }{ 2*4*6*...*(2n)}\right)^3=0$$ Having products at fractions I cant figure out how to calculate this limit.
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I think this will help: http://math.stackexchange.com/questions/1025630/to-show-for-following-sequence-lim-n-to-infty-a-n-0-where-a-n-1-3/1025669#1025669 – pointer Nov 25 '14 at 13:41
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Which limit do you want to compute? The one in the title or the one in the body of the question? – Rasmus Nov 25 '14 at 13:44
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the one in the body – GorillaApe Nov 25 '14 at 13:52
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Let $$a_n:=\frac{1\cdot3\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot\ldots\cdot(2n)} =\frac{(2n)!}{2^{2n}(n!)^2}$$
Then use the Stirling formula: $$ n! \sim \sqrt {2\pi n} n^n e^{-n} \\ a_n \sim \frac{\sqrt {4\pi n} (2n)^{2n} e^{-2n}} {2^{2n} 2\pi n n^{2n} e^{-2n}} = \frac{\sqrt {4\pi n} } {2\pi n } \to 0 $$
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