Conjecture: $\sum\limits_{n\geq0}\left(\frac12\right)^n\prod\limits_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}=\sqrt2$

The products can be rewritten as \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} \frac{1}{2^n} = \sum_{n=0}^{\infty} \binom{2n}{n} \frac{1}{8^n} \end{eqnarray*} Now use \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}. \end{eqnarray*}

Edit: \begin{eqnarray*} \prod_{k=1}^{n}\frac{2n-2k+1}{2n-2k+2}&=&\prod_{k=1}^{n}\frac{2k-1}{2k} =\frac{(2n-1)!!}{(2n)!!}\\ &=&\frac{(2n)!}{(2^n n!)^2}=\binom{2n}{n} \frac{1}{2^{2n}}. \end{eqnarray*}

Another way is $$ \eqalign{ & \prod\limits_{k = 1}^n {{{2n - 2k + 1} \over {2n - 2k + 2}}} = \prod\limits_{k = 1}^n {{{\left( {n - k} \right) + 1/2} \over {\left( {n - k} \right) + 1}}} = \prod\limits_{j = 0}^{n - 1} {{{j + 1/2} \over {j + 1}}} = \cr & = {{\left( {1/2} \right)^{\,\overline {\,n\,} } } \over {n!}} = {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} = \left( \matrix{ n - 1/2 \cr n \cr} \right) = \left( { - 1} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) \cr} $$ and then apply the binomial expansion.