Square Root of Wallis Product How can we prove or disprove this:
$$\begin{align}
\prod_{n=1}^{\infty}\frac {2n}{2n-1}=\frac {2\cdot 4\cdot 6\cdot 8\cdots}{1\cdot 3\cdot 5\cdot 7\cdots }
&=\sqrt2\cdot \sqrt{\frac{2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\
&=\sqrt{\frac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6\cdot8\cdots}{1\cdot3\cdot3\cdot5\cdot5\cdot7\cdot7\cdots}}\\
&=\sqrt{\frac{\pi}2}&&\text{(using Wallis Product)}
\end{align}$$
According to Wolframlpha, the product diverges, and the partial product is given by $$\prod_{n=1}^m\frac {2n}{2n-1}=\sqrt{\pi}\frac {\Gamma(m+1)}{\Gamma(m+\frac 12)}$$
The main objective of this question is ascertain why the Wallis product cannot be used in deriving the solution. 

Every terms count
\begin{align*}
  \frac{2 \cdot 4}{1\cdot 3} &= \sqrt{5} \times
  \sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4}{1\cdot 3 \cdot 3 \cdot 5}} \\
  \frac{2 \cdot 4 \cdot 6}{1\cdot 3 \cdot 5} &= \sqrt{7} \times
  \sqrt{\frac{2\cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6}
             {1\cdot 3 \cdot 3 \cdot 5 \cdot 5\cdot 7}} \\
  \frac{(2n))!!}{(2n-1)!!} &= \sqrt{2n+1} \times
  \sqrt{\prod_{k=1}^{n} \frac{4k^2}{4k^2-1}} \\
  & \sim \sqrt{n\pi}
\end{align*}
which blows up!!
