Double factorial series 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)?

 A: 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.
A: What is $n!!$ ? For $n=2k$, we have
$$ (2k)!! = 2k(2k-2)(2k-4)\dotsm (2), $$
which it is easy to see is
$$ 2^k k!. $$
Therefore the even sum is easy:
$$ \sum_{k=0}^{\infty} \frac{1}{(2k)!!} = \sum_{k=0}^{\infty} \frac{1}{k!}2^{-k} = e^{1/2}. $$
What about the odd terms?
$$ (2k+1)!! = (2k+1)(2k-1)\dotsm 3 \cdot 1 = \frac{(2k+1)!}{2^k k!} $$
This can be shown, by a fairly nasty process that I'll spare you, to be the following:
$$ \sum_{k=0}^{\infty} \frac{x^{2k+1}}{(2k+1)!!} = e^{x^2/2} \int_0^x e^{-t^2/2} \, dt. $$
After substituting $x=1$, you can't really get it into a more closed form than that (this is the same as the integral of the normal distribution's density function, which is well-known to not be expressible in closed form).
So
$$ \sum_{n=0}^{\infty} \frac{1}{n!!} = \sqrt{e} \left( 1 + \int_0^1 e^{-t^2/2} \, dt \right). $$
