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Possible Duplicate:
Another Evaluating Limit Question

What happens to the sequence $a_n=\frac{3\cdot 5\cdot7\cdot\ldots\cdot (2n-1)}{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}$ as $n$ tends to $\infty$? Would appreciate a sort of "proofish" thing. Thanks in advance.

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    Maybe [this](http://math.stackexchange.com/questions/936236/sum-of-the-series-frac12-cdot-4-frac1-cdot32-cdot4-cdot6-dots) is somewhat related2014-09-18

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You can write the sum as: $S=\prod_{k=1}^{N}\frac{2k+1}{2k}$ The result is $S=\frac{2(\frac{1}{2})^{N+1}2^{N+1}\Gamma(N+\frac{3}{2})}{\Gamma(N+1)\sqrt{\pi}}$

The limit of $S$ for $N\to \infty$ is $\infty$

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    That doesn't look right. You have $(\frac{1}{2})^{N+1}$ right next to $2^{N+1}$.2012-05-04