How to sum this series for $\pi/2$ directly?

First, $$(2k+1)!! = (2k+1)(2k-1) \cdots (1) = \frac{(2k+1)!}{(2k)(2(k-1)) \cdots 2(1)} = \frac{(2k+1)!}{2^k k!}.$$

So your sum can be rewritten as

$$\sum_{k=0}^\infty\frac{k! \, k! \, 2^k }{(2k+1)!} = \sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}.$$

Variations of the sum of reciprocals of the central binomial coefficients have been well-studied. For example, this paper by Sprugnoli (see Theorem 2.4) gives the ordinary generating function of $a_k = \frac{4^k}{(2k+1)}\binom{2k}{k}^{-1}$ to be $$A(t) = \frac{1}{t} \sqrt{\frac{t}{1-t}} \arctan \sqrt{\frac{t}{1-t}}.$$

Subbing in $t = 1/2$ says that $$\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}} = 2 \arctan(1) = \frac{\pi}{2}.$$

We can prove this identity, as well as the corresponding power series identities by using a relation with the Beta function. Rearranging as done in Mike Spivey's answer we are looking at $$ \sum_{k=0}^\infty\frac{k! k! 2^k}{(2k+1)!}$$ Using induction or a Beta Function identity, we can show that $$\int_0^1 x^{k}(1-x)^k=\frac{k!k!}{(2k+1)!}.$$ Hence your sum becomes

$$ \sum_{k=0}^\infty 2^k \int_0^1 x^{k}(1-x)^k=\int_0^1 \left(\sum_{k=0}^\infty 2^k x^k (1-x)^k\right)dx.$$

Notice that since $0\leq x\leq 1$, $x(1-x)\leq \frac{1}{4}$ and the series converges absolutely. Summing gives

$$=\int_0^1 \frac{1}{1-2x(1-x)}dx=\int_0^1 \frac{1}{x^2+(1-x)^2}dx$$ Substituting $u=\frac{1}{x}$, and then $v=u-1$, we see that this integral is equal to $$\int_1^\infty \frac{1}{1+(u-1)^2}du=\int_0^\infty \frac{1}{1+v^2}dv=\frac{\pi}{2},$$ as desired.

I had intended for this to be a comment to Mike Spivey's answer, but it is too long.

One of the answers to the related question mentions a result equivalent to $$ \int_0^\frac{\pi}{2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1} $$ Using $(1)$, my sum becomes $$ \begin{align} \sum_{k=0}^\infty\frac{k!}{(2k+1)!!} &=\sum_{k=0}^\infty\frac{2^k}{(2k+1)\binom{2k}{k}}\\ &=\sum_{k=0}^\infty\int_0^\frac{\pi}{2}\sqrt{2}\left(\frac{\sin(x)}{\sqrt{2}}\right)^{2k+1}\mathrm{d}x\\ &=\sqrt{2}\int_0^\frac{\pi}{2}\frac{\left(\frac{\sin(x)}{\sqrt{2}}\right)}{1-\left(\frac{\sin(x)}{\sqrt{2}}\right)^2}\;\mathrm{d}x\\ &=\int_0^\frac{\pi}{2}\frac{2\,\sin(x)}{2-\sin^2(x)}\;\mathrm{d}x\\ &=\int_\frac{\pi}{2}^0\frac{2\;\mathrm{d}\cos(x)}{1+\cos^2(x)}\\ &=\frac{\pi}{2} \end{align} $$