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

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

The product does not exhibit exponential decay. It goes down like the inverse of the square-root function. Plot it and you will see what I mean. — Batominovski, Aug 18 '18 at 14:12
isn't the inverse of the square root function the square function? — Lalaloopsy, Aug 18 '18 at 14:22
Yes, I meant the multiplicative inverse (i.e., the reciprocal). — Batominovski, Aug 18 '18 at 14:23
Alternatively: $$\prod_{k=1}^n,\left(\frac{2n-1}{2n}\right)=\frac{(2n-1)!!}{(2n+1)!!}=\frac{(2n)!}{4^n (n!)^2}$$ And now use https://en.wikipedia.org/wiki/Stirling%27s_approximation — Zacky, Aug 18 '18 at 14:30
@Zacky I have recently realized that Stirling's approximation and Wallis's product imply each other. If you know one, then you can prove the other. It's quite neat. — Batominovski, Aug 18 '18 at 14:49
@Zacky This is what I was talking about: https://math.stackexchange.com/questions/2078873/using-wallis-formula-to-show-the-limit-of-a-n-n-left-fracen-rightn-n. — Batominovski, Aug 18 '18 at 16:33

Batominovski · Accepted Answer · 2018-08-18T23:01:41.523

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.

Ahmad Bazzi · Answer 2 · 2018-08-18T15:06:49.767

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.

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

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