Finding $ \lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$ Finding 

$$\lim_{n \rightarrow \infty} \prod_{j=1}^n \frac{2j-1}{2j}$$

Suspecting that the limit is 0, but how do I show this? Was able to get an upper bound of $ \frac{1}{2\sqrt{e}} $ easily but it's not useful. Apparently this is a hard problem.

You may write
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots2n} &=\frac{1\cdot \color{blue}2\cdot 3\cdot \color{blue}4 \cdot 5\cdot\color{blue}6\cdots(2n-1)\cdot \color{blue}{2n}}{(2\cdot 4\cdot 6\cdots (2n))^\color{blue}2}\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2}\\
& =\frac{(2n)!}{2^{2n} (n!)^2 }\\
& =\frac{1}{\sqrt{\pi n}}+\mathcal{O}\left(\frac{1}{n^{3/2}}\right), \quad \text{for} \, n \, \text{great}
\end{align}
$$
where we have used Stirling's approximation (approximating $(2n)!$ and $n!$), then you easily conclude for the limit (here you have more than the limit).
