What happens to the sequence $a_n=\frac{3\cdot5\cdot7\cdot\ldots\cdot(2n-1)}{2\cdot4\cdot6\cdot\ldots\cdot2n}$ as $n$ tends to $\infty$? [duplicate]

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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marked as duplicate by Brian M. Scott, anon, J. M., t.b., Chandrasekhar May 4 '12 at 11:14

This question was marked as an exact duplicate of an existing question.

What happens to $b_n=\log(a_{n+1})-\log(a_n)$ when $n\to\infty$? – Did May 4 '12 at 11:03
The title of the duplicate is definitely better than the other title. – Did May 4 '12 at 11:24
Maybe this is somewhat related – Dinesh Sep 18 '14 at 13:55

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 $\infty$
That doesn't look right. You have $(\frac{1}{2})^{N+1}$ right next to $2^{N+1}$. – TonyK May 4 '12 at 11:15