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.
 A: 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).
A: Compare $A=\prod_{j=1}^n\frac{2j-1}{2j}$ with $B=\prod_{j=1}^n\frac{2j}{2j+1}$.
Each factor in A is less than the corresponding factor in B so $A<B$.
AB = $1/(2n+1)$ by the telescope principle.
$A^2<AB=1/(2n+1)$
$\lim_{n\to\infty}A=0$
A: I think that Stirling's approximation is an overkill. Notice that for any $j>1$:
$$\left(1-\frac{1}{2j}\right)^2 = 1-\frac{1}{j}+\frac{1}{4j^2} = \left(1-\frac{1}{j}\right)\cdot \left(1+\frac{1}{4j(j-1)}\right)\tag{1}, $$
so given that $P_N = \prod_{j=1}^{N}\left(1-\frac{1}{2j}\right)$, we have:
$$ P_N^2 = \frac{1}{4}\prod_{j=2}^{N}\left(1-\frac{1}{j}\right)\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right)=\frac{1}{4N}\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right).\tag{2}$$
Since $1+x<e^x$ for any $x\neq 0$, the last product can be bounded by:
$$\prod_{j=1}^{N-1}\left(1+\frac{1}{4j(j+1)}\right)\leq \exp\left(\frac{1}{4}\sum_{n=1}^{N-1}\frac{1}{n(n+1)}\right)\leq e^{\frac{1}{4}},\tag{3}$$
so we get the inequality:
$$ P_N \leq \frac{1}{2\sqrt{N}}e^{\frac{1}{8}} \tag{4} $$
that is enough to state $\lim_{N\to +\infty} P_N = 0$, as wanted. 

As an alternative approach, it is worth to recall that the infinite product
$$\prod(1+a_n)$$
is convergent iff $\sum a_n$ is convergent. Since 
$$\sum_{n\geq 1}\frac{1}{2n-1}$$
is divergent, the product
$$\prod_{n\geq 1}\left(\frac{2n}{2n-1}\right)$$
is divergent, too, so your product converges towards zero.
