$\displaystyle \lim_{n\to\infty}\frac{1\cdot 3\cdot 5 \cdot \cdots \cdot (2n-1)}{2\cdot 4\cdot 6 \cdot \cdots \cdot (2n)}$

i know the limit should be 0 because the the series in the numerator is larger the the one in the denominator for every n

but i don't know how to start on proving it

help?