Showing that $ \sum_{n=1}^{m} \prod_{k=1}^n \frac{2k-1}{2k} = \frac{2(m+1)\Gamma(m+\frac{3}{2})}{\sqrt{\pi}\Gamma(m+2)}$ I'm trying to show the following identity,
$$ \sum_{n=1}^{m} \prod_{k=1}^n \frac{2k-1}{2k} = \frac{2(m+1)\Gamma(m+\frac{3}{2})}{\sqrt{\pi}\Gamma(m+2)}.$$
First, simplify the product within the sum,
$$ \sum_{n=1}^{m} \frac{\prod_{k=1}^n (2k-1)}{\prod_{k=1}^n (2k)} \\
  = \sum_{n=1}^{m} \frac{\frac{(2n)!}{2^n n!}}{2^n n!} \\ 
 = \sum_{n=1}^{m} \frac{(2n)!}{2^{2n}(n!)^2} $$
I'm not sure where to continue from here. I know the factorials will be rewritten as the Gamma function and I'm guessing the $\sqrt{\pi}$ will come from Stirling's approximation. Where should I go from here?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{n = 1}^{m}\prod_{k = 1}^{n}{2k - 1 \over 2k} & =
\sum_{n = 1}^{m}{\prod_{k = 1}^{n}\pars{k - 1/2} \over \prod_{k = 1}^{n}k} =
\sum_{n = 1}^{m}{\pars{1/2}^{\overline{n}} \over n!} =
\sum_{n = 1}^{m}{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}
\\[5mm] & =
\sum_{n = 1}^{m}{\pars{n - 1/2}! \over n!\pars{-1/2}!} =
\sum_{n = 1}^{m}{n - 1/2 \choose n} =
\bracks{z^{m}}\sum_{\ell = 0}^{\infty}z^{\ell}
\bracks{\sum_{n = 1}^{\infty}{n - 1/2 \choose n}\bracks{n \leq \ell}}
\\[5mm] & =
\bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}
\sum_{\ell = n}^{\infty}z^{\ell} =
\bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}
\sum_{\ell = 0}^{\infty}z^{\ell + n}
\\[5mm] & =
\bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}z^{n} =
\bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-z}^{n} \\[5mm] & =
\bracks{z^{m}}{1 \over 1 - z}\bracks{\pars{1 - z}^{-1/2} - 1} =
\bracks{z^{m}}\pars{1 - z}^{-3/2} - \bracks{z^{m}}\pars{1 - z}^{-1}
\\[5mm] & =
{-3/2 \choose m}\pars{-1}^{m} - 1 =
{m + 1/2 \choose m} - 1 =
{\pars{m + 1/2}! \over m!\pars{1/2}!} - 1
\\[5mm] & =
{\Gamma\pars{m + 3/2} \over \Gamma\pars{m + 1}\Gamma\pars{3/2}} - 1 =
{\Gamma\pars{m + 3/2} \over
\bracks{\Gamma\pars{m + 2}/\pars{m + 1}}\bracks{\Gamma\pars{1/2}/2}} - 1
\\[5mm] & =
\bbx{{2\pars{m + 1}\Gamma\pars{m + 3/2} \over \root{\pi}\Gamma\pars{m + 2}} - 1}
\end{align}
A: Observe that from the RHS we have
$$\require{cancel}\begin{align}\frac{2(m+1)\Gamma(m+3/2)}{\sqrt\pi\Gamma(m+2)}&=\frac{2\cancel{(m+1)}(m+1/2)\Gamma(m+1/2)}{\Gamma(1/2)\cancel{(m+1)}\Gamma(m+1)}\\&=\frac{2(m+1/2)^\underline {m+1}\cancel{\Gamma(1/2)}}{m!\,\cancel{\Gamma(1/2)}}\\&=\binom{m+1/2}{m}=\prod_{k=1}^m\frac{k+1/2}{k}\\&=\prod_{k=1}^m\frac{2k+1}{2k}=\frac{(2m+1)!!}{(2m)!!}\tag{1}\end{align}$$
And from the LHS we have
$$\sum_{n=1}^{m} \frac{(2n)!}{2^{2n}(n!)^2}=\sum_{n=1}^{m}\frac{(2n)!}{((2n)!!)^2}=\sum_{n=1}^{m}\frac{(2n-1)!!}{(2n)!!}\tag{2}$$
To show that $(1)$ is equivalent to $(2)$ we can do this 
$$\Delta\left(\frac{(2n+1)!!}{(2n)!!}\right)=\frac{(2n+3)!!}{(2n+2)!!}-\frac{(2n+1)!!}{(2n)!!}=\left(\frac{2n+3}{2n+2}-1\right)\cdot\frac{(2n+1)!!}{(2n)!!}=\\=\frac1{2n+2}\cdot\frac{(2n+1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n+2)!!}$$
Then
$$\sum_{n=h}^{m-1}\frac{(2n+1)!!}{(2n+2)!!}=\sum_{n=h+1}^{m}\frac{(2n-1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n)!!}\bigg|_h^m\tag{3}$$
And from $(2)$ and $(3)$ we get
$$\sum_{n=1}^m\frac{(2n-1)!!}{(2n)!!}=\frac{(2n+1)!!}{(2n)!!}\bigg|_0^m=\frac{(2m+1)!!}{(2m)!!}-1$$
But there is a difference of $1$!! I double-checked the difference in Wolfram-Alpha, so you have a typo in your exercise.
A: $$
\begin{align}
\sum_{n=1}^m\prod_{k=1}^n\frac{2k-1}{2k}
&=\sum_{n=1}^m\binom{n-1/2}{n}\tag{1}\\
&=\sum_{n=1}^m(-1)^n\binom{-1/2}{n}\tag{2}\\
&=\sum_{n=1}^m(-1)^m\binom{-1}{m-n}\binom{-1/2}{n}\tag{3}\\
&=(-1)^m\binom{-3/2}{m}-1\tag{4}\\
&=\binom{m+1/2}{m}-1\tag{5}\\
&=\frac{\Gamma\left(m+\frac32\right)}{\Gamma(m+1)\Gamma\left(\frac32\right)}-1\tag{6}\\
&=\frac{\Gamma\left(m+\frac32\right)}{\frac1{m+1}\Gamma(m+2)\frac12\Gamma\left(\frac12\right)}-1\tag{7}\\
&=\frac{2(m+1)}{\sqrt\pi}\frac{\Gamma\left(m+\frac32\right)}{\Gamma(m+2)}-1\tag{8}
\end{align}
$$
Explanation:
$(1)$: convert product to Binomial Coefficient
$(2)$: convert to Negative Binomial Coefficient
$(3)$: rewrite $(-1)^n$
$(4)$: Vandermonde's Identity
$(5)$: convert from negative binomial coefficient
$(6)$: express binomial coefficient with Gamma functions
$(7)$: apply $x\Gamma(x)=\Gamma(x+1)$
$(8)$: $\Gamma\left(\frac12\right)=\sqrt\pi$
The sum would be as given in the question if the $n=0$ term were included in the sum.
