summation - 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?

Answer

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\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\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}