Wallis Product for $n = \tfrac{1}{2}$ From $n! = \Pi_{k=1}^\infty (\frac{k+1}{k})^n\frac{k}{k+n} $

Question

How does

$$ \prod_{k\ =\ 1}^{\infty}\left(\,\,\sqrt{\, k+1 \over k\,}\,{k \over k + 1/2}\,\right) ={\,\sqrt{\,\pi\,}\, \over 2} ={\sqrt{2\left(\,\pi/2\,\right)} \over 2} ={1 \over 2}\,\,\sqrt{\, 2\prod_{k\ =\ 1}^{\infty} {\left(\, 2k\,\right)^{2} \over \left(\, 2k - 1\,\right)\left(\, 2k + 1\,\right)}\,}\,\ {\large ?} $$

In other words, how do you derive the Wallis product ${\pi \over 2} =\prod_{k\ =\ 1}^{\infty} {\left(\, 2n\,\right)^{2} \over \left(\, 2n + 1\,\right)\left(\, 2n - 1\,\right)}$ from $\left(\,{1 \over 2}\,\right)! $ if $$ n! =\prod_{k\ =\ 1}^{\infty} \left[\,\left(\, k + 1 \over k\,\right)^{n}\,{k \over k+n}\,\right] $$

It should be possible, "after a bit of manipulation" but it never works!

Answer 1

From the Euler's identity: $$\begin{eqnarray*}z!=\Gamma(z+1) &=& \lim_{n\to +\infty}\frac{n! n^z}{(z+1)\cdot\ldots\cdot(z+n)}=\lim_{n\to +\infty}n^{z}\prod_{k=1}^{n}\left(1+\frac{z}{k}\right)^{-1}\\&=&\prod_{k=1}^{+\infty}\left(1+\frac{z}{k}\right)\left(1+\frac{1}{k}\right)^z\end{eqnarray*}$$ it follows that: $$\frac{1}{2}! = \prod_{k=1}^{+\infty}\sqrt{1+\frac{1}{k}}\left(1+\frac{1}{2k}\right)^{-1}$$ or: $$\left(\frac{1}{2}!\right)^2 = \prod_{k=1}^{+\infty}\left(1+\frac{1}{k}\right)\left(1+\frac{1}{k}+\frac{1}{4k^2}\right)^{-1}=\prod_{k=1}^{+\infty}\frac{(k+1)(2k)^2}{k(2k+1)^2}=\prod_{k=1}^{+\infty}\frac{(k+1)k^2}{k(k+1/2)^2}$$ that is just the square of our initial product. Since: $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},$$ it follows that $\Gamma(1/2)=\sqrt{\pi}$ and $$\frac{1}{2}!=\Gamma\left(\frac{3}{2}\right)=\color{red}{\frac{\sqrt{\pi}}{2}}$$ as wanted.

Answer 2

Let $$ F(n)=\prod_{k=1}^\infty\left(\frac{k+1}k\right)^n\frac{k}{k+n}\tag{1} $$ Then, using Stirling's Formula, $$ \begin{align} F\left(\frac12\right)^2 &=\prod_{k=1}^\infty\frac{k+1}k\frac{k^2}{\left(k+\frac12\right)^2}\\ &=\lim_{n\to\infty}\prod_{k=1}^n\frac{k+1}k\frac{4k^2}{(2k+1)^2}\\ &=\lim_{n\to\infty}(n+1)\left(\frac{4^n\color{#C00000}{n!^2}}{(2n+1)\color{#00A000}{(2n)!}}\right)^2\\ &=\lim_{n\to\infty}(n+1)\left(\frac{4^n\,\color{#C00000}{2\pi nn^{2n}e^{-2n}}}{(2n+1)\,\color{#00A000}{\sqrt{4\pi n}(2n)^{2n}e^{-2n}}}\right)^2\\ &=\lim_{n\to\infty}(n+1)\left(\frac{\sqrt{\pi n}}{2n+1}\right)^2\\[6pt] &=\frac\pi4\tag{2} \end{align} $$ Therefore, $$ F\left(\frac12\right)=\frac{\sqrt\pi}2\tag{3} $$

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Or Perhaps...

After I reread your question, it seems that perhaps you were looking for something like this: $$ \begin{align} \frac\pi4 &=\left(\tfrac12\right)!^2\tag{4a}\\ &=\prod_{k=1}^\infty\left(\sqrt{\frac{k+1}k}\frac{k}{k+\frac12}\right)^2\tag{4b}\\ &=\prod_{k=1}^\infty\frac{2k(2k+2)}{(2k+1)^2}\tag{4c}\\ &=\lim_{n\to\infty}\prod_{k=1}^n\frac{2k(2k+2)}{(2k+1)^2}\tag{4d}\\ &=\lim_{n\to\infty}\frac12\frac{2n+2}{2n+1}\prod_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)}\tag{4e}\\ &=\frac12\prod_{k=1}^\infty\frac{(2k)^2}{(2k-1)(2k+1)}\tag{4f} \end{align} $$ Explanation:
$\text{(4a)}$: use $(2)$
$\text{(4b)}$: identity given in question
$\text{(4c)}$: algebra
$\text{(4d)}$: definition of an infinite product
$\text{(4e)}$: $\left\{\begin{align} \textstyle\prod\limits_{k=1}^n\frac{2k(2k+2)}{(2k+1)^2} &\textstyle=\frac{2\cdot4}{3^2}\frac{4\cdot6}{5^2}\frac{6\cdot8}{7^2}\cdots\frac{(2n-2)2n}{(2n-1)^2}\frac{2n(2n+2)}{(2n+1)^2}\\ &\textstyle=\frac1{2^{\vphantom{1}}}\frac2{1^{\vphantom{1}}}\frac{2\cdot4}{3^2}\frac{4\cdot6}{5^2}\frac{6\cdot8}{7^2}\cdots\frac{(2n-2)2n}{(2n-1)^2}\frac{2n}{2n+1}\frac{2n+2}{2n+1}\\ &\textstyle=\frac{1^{\vphantom{1}}}2\underbrace{\frac{2^2}{1\cdot3}\frac{4^2}{3\cdot5}\frac{6^2}{5\cdot7}\cdots\frac{(2n)^2}{(2n-1)(2n+1)}}\frac{2n+2}{2n+1}\\ &\textstyle=\frac12\frac{2n+2}{2n+1}\prod\limits_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)} \end{align}\right.$
$\text{(4f)}$: $\lim\limits_{n\to\infty}\frac{2n+2}{2n+1}=1$ and the definition of an infinite product

Therefore, $$ \frac\pi2=\prod_{k=1}^\infty\frac{(2k)^2}{(2k-1)(2k+1)}\tag{5} $$