Wallis this?

We begin by simplifying the sum:

$\begin{aligned} \sum_{m=1}^\infty \frac{(2m-1)!!}{(2m)!!} \left( \frac{7}{16} \right)^m &= \sum_{m=1}^\infty \frac{(2m)!/[m! 2^m]}{m! 2^m} \left( \frac{7}{16} \right)^m \\ &= \sum_{m=1}^\infty \binom{2m}{m} \left( \frac{7}{64} \right)^m \\ &= -1 + \sum_{m=0}^\infty \binom{2m}{m} \left( \frac{7}{64} \right)^m. \end{aligned}$

Note that the generating function for $\dbinom{2m}{m}$ is $\displaystyle \sum_{m=0}^\infty \binom{2m}{m} x^m = \frac{1}{\sqrt{1-4x}}$ . Thus,

$\sum_{m=0}^\infty \binom{2m}{m} \left( \frac{7}{64} \right)^m = \frac{1}{\sqrt{1-4(7/64)}} = \frac{4}{3}.$

Therefore, $\displaystyle \sum_{m=1}^\infty \frac{(2m-1)!!}{(2m)!!} \left( \frac{7}{16} \right)^m = -1+\frac{4}{3} = \boxed{\dfrac{1}{3}}$ .