Limit of the Product $\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)$

Question

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

$$\lim_{n\rightarrow\infty}\left(1+\frac{-1/2}{n}\right)^n$$

Answer 1

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.

Answer 2

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.