<s>[INST] Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge? 
Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?

This is a homework question for an analysis class.
Firstly, I know the sequence is comprised of odd numbers up to $2n-1$ over even numbers up to $2n$.
Secondly, I know the sequence decreases. I'm guessing it'll converge, but not specifically to $0$.
Where I am stuck is simplifying the series to something easier to work with. Any help would be greatly appreciated!
Edit: solved, thank you everyone. I ended up doing what was first suggested, not finding the actual limit but just proving the sequence converges. Thanks again!
 [/INST] $$a^{ 2 }_{ n }={ \left( \frac { 1\cdot 3\cdot 5\cdot ...\cdot (2n-1) }{ 2\cdot 4\cdot 6\cdot ...\cdot (2n) }  \right)  }^{ 2 }=\frac { { 1 }^{ 2 }\cdot 3^{ 2 }\cdot 5^{ 2 }\cdot ...\cdot (2n-1)^{ 2 } }{ 2^{ 2 }\cdot 4^{ 2 }\cdot 6^{ 2 }\cdot ...\cdot (2n)^{ 2 } } =\\ =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\cdot 5 }{ 4^{ 2 } } \cdot \frac { \left( 2n-1 \right) \left( 2n+1 \right)  }{ \left( 2n \right) ^{ 2 } } \cdot \frac { 1 }{ 2n+1 } <\frac { 1 }{ 2n+1 } \\ \\ $$
for every $n$ $  \epsilon$ $  \mathbb{N}   $ 
$0<{ a }_{ n }<\frac { 1 }{ \sqrt { 2n+1 }  } \quad $ so it converges and

$$ \lim _{ n\rightarrow \infty  }{ { a }_{ n } } =0$$

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