Find the value of $(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots$. Show that: $$(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots = \dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}.$$
By the duplication theorem and $\Gamma(z+1)=z\,\Gamma(z)$, we have: $$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=\dfrac{\Gamma\left(-\dfrac{z}{2}\right)}{z\,\Gamma\left(-z\right)\Gamma\left(\dfrac{z}{2}\right)} \cdot 2^{-z}.$$
Using the definition $\Gamma\left(z\right) = \dfrac{1}{z}\prod_{k=1}^\infty \left(1+\dfrac{1}{k}\right)^z \left(1+\dfrac{z}{k}\right)^{-1}$, and simplifying terms, we obtain:$$\dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}=2^{-z}(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots .$$
How can I get rid of the $2^{-z}$, and am I going about this the right way?
Cheers!
 A: We can make use of Euler's product for the gamma function, Whittaker and Watson, page 237, §12.11, also (31) of MathWorld:
$$
\frac{1}{\Gamma(z)} =
\lim_{n\rightarrow\infty}
\frac{z \, (1+ z) \cdots (n-1 + z)}{1 \cdot 2 \cdots (n - 1)} n^{-z}.
\qquad (1)
$$
According to (1), we have
$$
\begin{align}
\frac{1}{\Gamma(1 + \frac z 2)}
&=
\lim_{n\rightarrow\infty}
\frac{(1 + \frac{z}{2}) \, (2+\frac{z}{2}) \cdots (n + \frac{z}{2})}{1 \cdot 2 \cdots (n - 1)} n^{-1 - z/2} \\
&=
\lim_{n\rightarrow\infty}
\left(1 + \frac{z}{2}\right) \,
\left(1 + \frac{z}{4}\right) \cdots \left(1 + \frac{z}{2n}\right) n^{-z/2}.
\qquad (2)
\end{align}
$$
Similarly,
$$
\begin{align}
\frac{1}{\Gamma(\frac 1 2 - \frac z 2)}
&=
\lim_{n\rightarrow\infty}
\frac{(\frac{1}2 - \frac{z}{2}) \, (\frac{3}{2} -\frac{z}{2}) \cdots (\frac{2n - 1}{2} -\frac{z}{2})}{1 \cdot 2 \cdots (n - 1)} n^{-1/2 + z/2}
\\
&=
\lim_{n\rightarrow\infty}
\frac{1}{2} \frac{1 \cdot 3 \cdots (2n - 1)}{2 \cdot 4 \cdots (2n-2) \sqrt{n} }
\left(1 - z\right) \,
\left(1 - \frac{z}{3}\right) \cdots \left(1 - \frac{z}{2n-1}\right) n^{z/2},
\\
&=
\lim_{n\rightarrow\infty}
\frac{1}{\sqrt{\pi}} \,
\left(1 - z\right) \,
\left(1 - \frac{z}{3}\right) \cdots \left(1 - \frac{z}{2n-1}\right) n^{z/2},
\qquad (3)
\end{align}
$$
where in the last step we have used  Wallis's product formula:
$$
\lim_{n\rightarrow\infty}
\left(
\frac{1}{2} \frac{1 \cdot 3 \cdots (2n - 1)}{2 \cdot 4 \cdots (2n-2) \sqrt{n} }
\right)^2
=
\lim_{n\rightarrow\infty}
\frac{1}{2} \frac{1 \cdot 3 \cdot 3 \cdot 5 \cdots \cdot (2n - 1) \cdot (2n - 1)}
{2 \cdot 2 \cdot 4 \cdot 4 \cdots \cdot (2n - 2) \cdot (2n) }
= \frac{1}{2} \times \frac{2}{\pi}  = \frac{1}{\pi}.
$$
Multiplying the two yields
$$
\begin{align}
\frac{1}{\Gamma(1 + \frac z 2)\,\Gamma(\frac 1 2 - \frac z 2)}
&=
\frac{1}{\sqrt\pi}
\left(1 - z\right) \,
\left(1 + \frac{z}{2}\right) \,
\left(1 - \frac{z}{3}\right) \,
\left(1 + \frac{z}{4}\right) \,
\cdots,
\end{align}
$$
which is the desired result.
A: I would do the following: consider a finite product
$$\mathcal{P}_N:=(1-z)\left(1+\frac z{2}\right)\ldots \left(1-\frac{z}{2N-1}\right)\left(1+\frac{z}{2N}\right)$$
and rewrite it as
\begin{align*}\mathcal{P}_N&=\frac{\prod_{k=0}^{N-1}\left(2k+1-z\right)\prod_{k=0}^{N-1}\left(2k+2+z\right)}{(2N)!}=\\
&=\frac{2^{2N}}{(2N)!}{\color{red}{\prod_{k=0}^{N-1}\left(k+\frac{1-z}{2}\right)}}{\color{blue}{
\prod_{k=0}^{N-1}\left(k+1+\frac{z}{2}\right)}}=\\&=
\frac{2^{2N}}{(2N)!}{\color{red}{\frac{\Gamma\left(N+\frac{1-z}{2}\right)}{
\Gamma\left(\frac{1-z}{2}\right)}}}{\color{blue}{\frac{\Gamma\left(N+1+\frac{z}{2}\right)}{
\Gamma\left(1+\frac{z}{2}\right)}}}=\\
&=\frac1{\Gamma\left(\frac{1-z}{2}\right)\Gamma\left(1+\frac{z}{2}\right)}
\times\underbrace{
\frac{2^{2N}\Gamma\left(N+\frac{1-z}{2}\right)\Gamma\left(N+1+\frac{z}{2}\right)}{\Gamma(1+2N)}}_{\mathcal{Q}_N}.
\end{align*}
Now it suffices to use Stirling formula to show that $\mathcal{Q}_{\infty}=\sqrt\pi$. Indeed, as $N\to\infty$, one has
\begin{align*}
\mathcal Q_N&= 2^{2N}\frac{\sqrt{\frac{2\pi}{N+\frac{1-z}{2}}}\left(\frac{N+\frac{1-z}{2}}{e}\right)^{N+\frac{1-z}{2}}\cdot \sqrt{\frac{2\pi}{N+1+\frac{z}{2}}}\left(\frac{N+1+\frac{z}{2}}{e}\right)^{N+1+\frac{z}{2}}}{\sqrt{\frac{2\pi}{2N+1}}\left(\frac{2N+1}{e}\right)^{2N+1}}\left[1+O\left(N^{-1}\right)\right]=\\
&= \sqrt{\pi}\cdot  e^{-\frac12}
\frac{\left(1+\frac{1-z}{2N}\right)^N\left(1+\frac{1+\frac z2}{N}\right)^N}{\left(1+\frac1{2N}\right)^{2N}}\left[1+O\left(N^{-1}\right)\right]=
\sqrt{\pi}\left[1+O\left(N^{-1}\right)\right].
\end{align*}
