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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$$\begin{align} \sum_{n = 1}^{m}\prod_{k = 1}^{n}{2k - 1 \over 2k} & = \sum_{n = 1}^{m}{\prod_{k = 1}^{n}\pars{k - 1/2} \over \prod_{k = 1}^{n}k} = \sum_{n = 1}^{m}{\pars{1/2}^{\overline{n}} \over n!} = \sum_{n = 1}^{m}{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!} \\[5mm] & = \sum_{n = 1}^{m}{\pars{n - 1/2}! \over n!\pars{-1/2}!} = \sum_{n = 1}^{m}{n - 1/2 \choose n} = \bracks{z^{m}}\sum_{\ell = 0}^{\infty}z^{\ell} \bracks{\sum_{n = 1}^{\infty}{n - 1/2 \choose n}\bracks{n \leq \ell}} \\[5mm] & = \bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n} \sum_{\ell = n}^{\infty}z^{\ell} = \bracks{z^{m}}\sum_{n = 1}^{\infty}{n - 1/2 \choose n} \sum_{\ell = 0}^{\infty}z^{\ell + n} \\[5mm] & = \bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{n - 1/2 \choose n}z^{n} = \bracks{z^{m}}{1 \over 1 - z}\sum_{n = 1}^{\infty}{-1/2 \choose n}\pars{-z}^{n} \\[5mm] & = \bracks{z^{m}}{1 \over 1 - z}\bracks{\pars{1 - z}^{-1/2} - 1} = \bracks{z^{m}}\pars{1 - z}^{-3/2} - \bracks{z^{m}}\pars{1 - z}^{-1} \\[5mm] & = {-3/2 \choose m}\pars{-1}^{m} - 1 = {m + 1/2 \choose m} - 1 = {\pars{m + 1/2}! \over m!\pars{1/2}!} - 1 \\[5mm] & = {\Gamma\pars{m + 3/2} \over \Gamma\pars{m + 1}\Gamma\pars{3/2}} - 1 = {\Gamma\pars{m + 3/2} \over \bracks{\Gamma\pars{m + 2}/\pars{m + 1}}\bracks{\Gamma\pars{1/2}/2}} - 1 \\[5mm] & = \bbx{{2\pars{m + 1}\Gamma\pars{m + 3/2} \over \root{\pi}\Gamma\pars{m + 2}} - 1} \end{align}$$