TITLE: Double factorial series
QUESTION [5 upvotes]: My question is pretty simple.
Since $n! \gt n!!$, it's clear by the comparison test that $\sum_{n=0}^\infty \frac {1}{n!!}$ converges. 

But what value does the sum converge to? How does one go about determining its value (if possible)?

REPLY [4 votes]: I'd like to add that the "nasty process" mentioned by Chappers is indeed not that nasty. You can take $\varphi(x)=\mathrm{e}^{x^2/2}\int_0^x\mathrm{e}^{-t^2/2}\,\mathrm{d}t$ which is a differentiable function whose derivative equals $2x\varphi(x)+1$, so it is actually a $C^\infty$ function. Then, by applying Leibniz' rule we can see that the odd order derivatives of $\varphi$ satisfy $$\varphi^{(2k+1)}(x)=2x\varphi^{(2k)}(x)+2k\varphi^{(2k-1)}(x)$$ from where $\varphi^{(2k+1)}(0)=2^kk!=(2k)!!$. Furthermore, since $\varphi$ is an odd function, its even order derivatives all vanish at $0$ and so $$\varphi(x)=\sum_{k=0}^\infty\frac{(2k)!!}{(2k+1)!}x^k.$$ Since $(2k+1)!!(2k)!!=(2k+1)!$ we can conclude the equality.