Limit of the Product $\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)$ I know I can transform the above product as follows
\begin{eqnarray*}\lim_{n\rightarrow\infty}\left(\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}\right)&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)\\&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\&=&\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right)
\end{eqnarray*}
My thoughts are this is going to go to $e^{-x/2}$, but I can't figure out how to go from 
$$\lim_{n\rightarrow\infty}\prod_{k=1}^n\left(1+\frac{-1/2}{k}\right)$$
to
$$\lim_{n\rightarrow\infty}\left(1+\frac{-1/2}{n}\right)^n$$
 A: From Wallis's Product Formula
$$\prod_{n=1}^\infty\,\left(\frac{2n-1}{2n}\right)\,\left(\frac{2n+1}{2n}\right)=\frac2\pi\,,$$
we see that
$$(2m+1)\,\left(\prod_{n=1}^m\,\left(\frac{2n-1}{2n}\right)\right)^2=\prod_{n=1}^m\,\left(\frac{2n-1}{2n}\right)\,\left(\frac{2n+1}{2n}\right)\approx \frac{2}{\pi}\,.$$
Consequently,
$$\prod_{n=1}^m\,\left(\frac{2n-1}{2n}\right)\approx\sqrt{\frac{2}{\pi(2m+1)}}\,.$$
For comparison plots, see here and here.
A: If you want to prove that the limit goes to zero using exponentials, as you mention in your question, then take the $\ln$, let
$$a_n = \prod_{k=1}^n\left(\frac{2k-1}{2k}\right)$$ 
then call
$$b_n = \ln a_n = \ln \prod_{k=1}^n\left(\frac{2k-1}{2k}\right) = \sum\limits_{k=1}^n \ln \big( 1 - \frac{1}{2k} \big) $$ We know that $\ln(1-x) \leq -x$. So 
$$b_n \leq -\frac{1}{2} \sum\limits_{k=1}^n \frac{1}{k}$$
Now let $c_n = \sum\limits_{k=1}^n \frac{1}{k}$, which is the harmonic series and we know that $\lim_{n \rightarrow \infty} c_n = + \infty$. Take exponentials on both sides of the above equation, we get
$$a_n \leq e^{-\frac{1}{2} c_n}$$
Now, the limit is 
$$0 \leq \lim a_n \leq \lim e^{-\frac{1}{2} c_n} = e^{-\infty} = 0$$
So, by the Sandwich theorem,  the limit goes to zero. 
NOTE: @Batominovski gives another approach on how the product behaves asymptotically, i.e. as $\prod_{n=1}^m\,\left(\frac{2n-1}{2n}\right)\approx\sqrt{\frac{2}{\pi(2m+1)}}\,.$, which (in my opinion) is more interesting.
