Step 1: Recognise the binomial pattern.
The product $1\cdot3\cdot5\cdots(2r-1)$ in the numerator signals the expansion of $(1-x)^{-1/2}$, whose general term is $\dfrac{1\cdot3\cdot5\cdots(2r-1)}{2^r\, r!}x^r$.
Step 2: Reshape the given series.
Our series has $2^{2r} = 4^r$ in the denominator and $(\sqrt3)^r$ on top. Split $4^r = 2^r\cdot 2^r$, so \[ S = \sum_{r=1}^{\infty}\frac{1\cdot3\cdots(2r-1)}{2^r\,r!}\left(\frac{\sqrt3}{2}\right)^r. \]
Step 3: Identify $x$.
This matches $(1-x)^{-1/2} - 1$ with $x = \dfrac{\sqrt3}{2}$ (the $-1$ removes the $r=0$ term).
Step 4: Sum the series.
Therefore \[ S = \left(1 - \frac{\sqrt3}{2}\right)^{-1/2} - 1 = \sqrt{\frac{2}{2-\sqrt3}} - 1. \]
Step 5: Rationalise inside the root.
Multiply by $\dfrac{2+\sqrt3}{2+\sqrt3}$: since $(2-\sqrt3)(2+\sqrt3) = 1$, we get $\sqrt{2(2+\sqrt3)} = \sqrt{4+2\sqrt3} = \sqrt{(\sqrt3+1)^2} = \sqrt3 + 1$, so $S = \sqrt3$.
Step 6: Match the option.
The same value is written as $\sqrt{\dfrac{3}{\sqrt3-1}} - 1$, since $\sqrt{\dfrac{3}{\sqrt3-1}} = \sqrt3 + 1$ as well. This is option (C).
\[ \boxed{\sqrt{\dfrac{3}{\sqrt3-1}} - 1} \]